visual c++ samples and examples

Sparse Matrix

2010-11-25T00:02:00.000-08:00

Sparse Matrix

Numerical analysis in which a matrix most elements are zeros

Introduction
in some situation we face with matrix that they have a lot of zeros element and as you know it is not necessary to multiple,add,... all of them.if there exists a way to ignore them then the matrix algorithms go better and more efficient.
so we can store matrix as list and everywhere that the element is zero we dont store it but in referencing to it we return zero.
if you do a quick look at it you can see the algorithm,and if you run it you will enjoy

//in the name of allah
#include
using namespace std;
class sparseElem{
public:
    int row,col;
    double value;
    sparseElem *nextRow;
    sparseElem *nextCol;
    sparseElem(){
    }

};

class sparseMatrix{
public:
    int n,m;
    sparseElem *head_col;
    sparseElem *head_row;
    sparseElem*c;
    sparseElem *r;
    sparseMatrix(){
       m=3;n=3;
      head_col=head_row=NULL;
      c=new sparseElem[n];
      r=new sparseElem[m];
      for(int i=0;i
        r[i].row=i;
      for(int i=0;i
        c[i].col=i;
     }
    sparseMatrix(int m1,int n1){
      m=m1;n=n1;
      head_col=head_row=NULL;
      c=new sparseElem[n];
      r=new sparseElem[m];
      for(int i=0;i
        r[i].row=i;
      for(int i=0;i
        c[i].col=i;
    }
    void setElem(int i,int j,double x){
      sparseElem *tmp=new sparseElem;
        tmp->row=i;
        tmp->col=j;
        tmp->value=x;
        //important
        tmp->nextRow=NULL;                                        //point
        tmp->nextCol=NULL;
      if(!head_row){//head_row=head_col=NULL
        head_row=r+i;
        head_col=c+j;
        r[i].nextRow=tmp;
        r[i].nextCol=NULL;
        c[j].nextCol=tmp;
        c[j].nextRow=NULL;
        tmp->nextRow=NULL;
        tmp->nextCol=NULL;
      }
      else{
        sparseElem *pc=head_col;
        sparseElem *lpc=NULL;
        while(pc->nextRow!=NULL && tmp->col>pc->col){
        //cout<<"while 1"<
          lpc=pc;
          pc=pc->nextRow;
        }
        if(tmp->col < pc->col){
          if(lpc)
            lpc->nextRow=c+j;
          else
            head_col=c+j;
          (c+j)->nextRow=pc;
          c[j].nextCol=NULL;                                                                    //new
        }

        else if(tmp->col > pc->col){
          pc->nextRow=c+j;
          c[j].nextRow=NULL;                                      //new added                                                         maybe error
          c[j].nextCol=NULL;                                            //new added maybe optional
        }
        else//both are equal
          pc=c+j;
        //fixing r[i] ha
        sparseElem *pr=head_row;
        sparseElem *lpr=NULL;
        while(pr->nextCol!=NULL && tmp->row > pr->row){
        //cout<<"while 2"<
          lpr=pr;
          pr=pr->nextCol;
        }
        if(tmp->row < pr->row){
          if(lpr)
            lpr->nextCol=r+i;
          else
            head_row=r+i;//the first element is inserting;
          (r+i)->nextCol=pr;
          r[i].nextRow=NULL;                                                                //new
        }
        else if(tmp->row>pr->row){
          pr->nextCol=r+i;
          r[i].nextCol=NULL;                                      //new added
          r[i].nextRow=NULL;//r[i].nextCol=NULL;
        }
        else//both are equal
          pr=r+i;
        //fixing adding element to a others in sparseMatrix

        sparseElem *pc2=c+j;
        sparseElem *lpc2=NULL;
        while(pc2->nextCol!=NULL &&pc2->nextCol->rowrow){
           lpc2=pc2;
           pc2=pc2->nextCol;//chon dar haman sotun search mikonim
        }
        //agar az ghabk vojud dash faghat value ra be an midahim vagarne hame chiz be ham mirizad va hafeze ezafe mibarad
        if(pc2->nextCol && pc2->nextCol->row ==tmp->row)//      ! if(pc2->nextCol && pc2->nextCol->row>tmp->row)//                   new   error if(pc2->nextCol )
          pc2->nextCol->value=tmp->value;
          //tmp=NULL;//delete tmp; next is deleting it in below
        else {//if(pc2->nextCol && pc2->nextCol->row > tmp->row){//``````````````````completely new
          lpc2=pc2;
          tmp->nextCol=lpc2->nextCol;
          pc2->nextCol=tmp;
          }
        sparseElem *pr2=r+i;
        sparseElem *lpr2=NULL;
        while(pr2->nextRow!=NULL &&pr2->nextRow->colcol){
           lpr2=pr2;
           pr2=pr2->nextRow;//chon dar haman sotun search mikonim
        }
        if(pr2->nextCol && pr2->nextRow->col ==tmp->col){
          delete tmp;
          }
        else{// if(pc2->nextCol && pc2->nextCol->col > tmp->col){
          lpr2=pr2;                        //       new error //lpr2=lpr;
          tmp->nextRow=lpr2->nextRow;
          pr2->nextRow=tmp;
        }
      }
    }
    double getElem(int i,int j){
      sparseElem *pc=c+j;
      if(pc==NULL)//if there is no element in matrix returns 0
        return 0;
      while(pc->nextCol!=NULL){
        pc=pc->nextCol;
        if(pc->row==i)
          return pc->value;
        else if(pc->row>i)
          return 0;//if the row has been passed
      }
      return 0;
    }

    sparseMatrix *sparseAdd(sparseMatrix &A,sparseMatrix &B){
      sparseMatrix* C=new sparseMatrix(m,n);
      sparseElem *apr=A.head_row;
      while(apr!=NULL){
          sparseElem* ap=apr->nextRow;
          while(ap){//p!=null
            C->setElem(ap->row,ap->col,A.getElem(ap->row,ap->col));
            ap=ap->nextRow;
          }
          apr=apr->nextCol;
      }
      double value;
      sparseElem *bpr=B.head_row;
      while(bpr!=NULL){
          sparseElem* bp=bpr->nextRow;
          while(bp){//p!=null
            value=C->getElem(bp->row,bp->col)+B.getElem(bp->row,bp->col);
            C->setElem(bp->row,bp->col,value);
            bp=bp->nextRow;
          }
          bpr=bpr->nextCol;
      }
     return C;
    }
    sparseMatrix *sparseMult(sparseMatrix &A,sparseMatrix &B){
      sparseMatrix *C=new sparseMatrix(A.m,A.n);
      sparseElem *ar=A.head_row;

      while(ar){
        sparseElem *r=ar->nextRow;
        sparseElem *bc=B.head_col;
        while(bc){
            //bc=bc->nextRow;
            sparseElem *c=bc->nextCol;
            double value=0;
            while(c && r && c->rowcol)
                c=c->nextCol;
            while(r && c && r->colrow)
                r=r->nextRow;
            while(r && c){
              value+=r->value*c->value;
              r=r->nextRow;
              c=c->nextCol;
            }
            if(value)
              C->setElem(ar->row,bc->col,value);
            bc=bc->nextRow;//indexing
            r=ar->nextRow;
        }
        ar=ar->nextCol;
      }
      return C;
    }
};
int main(){
int m,n;
cout<<"please enter m,n:"<
cin>>m>>n;
sparseMatrix M(m,n);
sparseMatrix N(m,n);
double a;

cout<<"please enter two matrix:"<
for(int i=0;i
    for(int j=0;j
      cin>>a;
      if(a){
        M.setElem(i,j,a);
      }
    }
cout<
for(int i=0;i
    for(int j=0;j
      cin>>a;
      if(a){
        N.setElem(i,j,a);
      }
    }
sparseMatrix *C=M.sparseMult(M,N);
cout<<"this is multiplaying two matrices:"<
for(int i=0;i
    for(int j=0;j
      a=C->getElem(i,j);
      cout<<<" ";
    }
cout<
}

}

Heap sorting Class

2010-11-07T11:06:00.000-08:00

Heap Sorting class
template T& PQueue::min(){
return heap[0];
}
template T& PQueue ::extractMin(){
//cout<<"12"<
if(!heapsize)throw runtime_error(" the priority queue is empty\n");
T *min=new T;
//cout<<"13"<
min[0]=heap[0];
//cout<<"14"<
//cout<<"heapsize:"<<
swap(heap[0],heap[--heapsize]);//heap[0]=heap[heapsize--];
//cout<<"15"<
minHeapify(0);
//cout<<"16"<
return min[0];
}
templatevoid PQueue::minHeapify(int i){
int l=i*2+1;//important indexing from zero
int r=(i+1)*2;//important indexing from zero
int lowest;
//cout<<"17"<
if(lheap[l]){lowest=l;}
else {lowest=i;}
if(rheap[r]){lowest=r;}
if(lowest!=i){
    //cout<<"21 lowest="<<
    //cout<<<" heap["<<<"]="<<
    //cout<<<<"before tree is:"<
    //print();
    //cout<
    swap(heap[lowest],heap[i]);
    //cout<<"after swap:"<<<" heap["<<<"]="<<
    //cout<<<<"after tree is:"<
    //print();
    //cout<
    //cout<<"22"<
    minHeapify(lowest);
    //cout<<"23"<
}
}
template void PQueue::swap(T &p,T &q){
T tmp=p;//code_h tmp
p=q;
q=tmp;
}
template void PQueue::Delete(int e){
if(!heapsize)throw runtime_error(" the priority queue is empty\n");
T *min=new T;
swap(heap[e],heap[--heapsize]);//if indexing from 1 for heapsize then:heap[0]=heap[heapsize--];
minHeapify(e);
}
templatevoid PQueue::increaseKey(int e){
heap[e]++;
minHeapify(e);
}
templatevoid PQueue::decreaseKey(int e){
heap[e]--;
minHeapify(e);
}
templatevoid PQueue::insert(T &e){
//cout<<"43"<
if(heapsize>=n) throw runtime_error("invalid address :you are inserting more than n(size)\n");
heap[heapsize]=e;//heapsize begin from 0
buildheap(heapsize++);
}
templatevoid PQueue::insert(T *e){
//cout<<"53"<
if(heapsize>=n) throw runtime_error("invalid address :you are inserting more than n(size)\n");
heap[heapsize]=*e;//heapsize begin from 0
//cout<<"60"<
buildheap(heapsize++);
}
templatevoid PQueue::buildheap(int h){
while(h>0 && heap[h] s=(k-1)/2
      swap(heap[(h-1)/2],heap[h]);
      h=(h-1)/2;
}
}
templatebool PQueue::empty() const{
return heap==heap+heapsize;
}
templatevoid PQueue::print() const{
int i=0;
cout<<"the elements in the PriorityQueue:"<
while(i
    cout<<
    i++;
}
}

Huffman encoding and decoding using template

2010-11-04T13:05:00.000-07:00

Huffman encoding and decoding   using template

This Hufffman using encoding and decoding of characters through iterating   within a file of text

//in the name of allah
#include
#include
#include
#include"code_char.h"
#include"heap_class.h"
#include"heap.h"

using namespace std;
int paus;
bool read_file(char* en_words,int *number,list *w){//this template must be used when we add to heap ,PQueue* heap(int * a)=0,int * number=0){
fstream file;
file.open(en_words,ios::in);
if(!file)
    return false;
//code_ch *w=new code_ch[200];
file.seekg(0,ios::end);
int end=file.tellg();
file.seekg(0,ios::beg);
int i=0;
char ch;
while(i
    file.seekg(i,ios::beg);
    file.read((char*)&ch,1);
    i++;
    //if(ch=='\n' || ch=='\13')
      //continue;
//       if(ch=='\13')
//     continue;
    if(!number[ch]){//if is not yet inserted to list

      code_ch *tmp=new code_ch;
      number[ch]++;
      tmp->ch=ch;
      tmp->number=number[ch];//later we update this
      w->push_back(*tmp);//w->push_front(tmp);///
    }
    else
      number[ch]++;
}
//updateing the numbers
list::iterator p=w->begin();
while(p!=w->end()){                                                                             ///point
    p->number=number[p->ch];
    p++;
}
//copy(w->begin(),w->end(),ostream_iterator(cout,""));
return true;
}
code_ch * huffman(PQueue &p,int n){
code_ch *x;//=new code_ch;
code_ch *y;//=new code_ch;
for(int i=1;i
    code_ch *z=new code_ch;
    z->left=x=&p.extractMin();
    z->right=y=&p.extractMin();
    z->number=x->number+y->number;
    p.insert(z);
}
return &p.extractMin();
}
//this function finds the code of eeach element in the huffman tree that i create.
void find_code(code_ch *h,int i=-1){///h is root
i++;
if(h->left!=NULL){
    for(int k=0;k
      if(h->code[0][k])
    h->left->code[0][k]=true;
      else
    h->left->code[0][k]=false;
    }
    h->left->code[0][i]=false;
    h->left->code_length=h->code_length+1;
    find_code(h->left,i);
}
if(h->right!=NULL){
    for(int k=0;k
      if(h->code[0][k])
    h->right->code[0][k]=true;
      else
    h->right->code[0][k]=false;
    }
    h->right->code[0][i]=true;
    h->right->code_length=h->code_length+1;
    find_code(h->right,i);
}
}
//this function "mirizad" tree in     an array for useing when encodeing from file for each character we can find the element in the tree easily.
code_ch ** inar(code_ch *root){//,int tree_size){
static code_ch **l=new code_ch*[200];//l[197] points to an element in the tree that its character is 'a'
l[root->ch]=root;
if(root->left!=NULL)
     inar(root->left);//,tree_size);
if(root->right!=NULL)
     inar(root->right);//,tree_size);
return l;
}
int get_size(code_ch *root){//returns number of bits
   static int size=0;
   if(root->left!=NULL)
     get_size(root->left);
   if(root->right!=NULL)
     get_size(root->right);
   if(!root->left || !root->right)
     size+=root->number*root->code_length;
   return size;//this is number of bits
}
void encodeing(code_ch *root,char * en_file,code_ch **Arcode,int size){
//cout<<"coming in incodeing"<
fstream rfile,wfile;
rfile.open(en_file,ios::in);
wfile.open("encoded.txt",ios::out);
if(!rfile || !wfile)
     throw runtime_error("cannot open the files...\n");
rfile.seekg(0,ios::end);
int end=rfile.tellg();
rfile.seekg(0,ios::beg);
int i=0,seek=0,j=0;
char ch;
//cout<<"declaring A"<
bitArray A(size);
//cout<<"this is size="<<
//cout<<"after"<
while(i
    rfile.seekg(i,ios::beg);
    rfile.read((char*)&ch,1);
    //cout<<<" ";
    //cout<<" "<
    for(j=0;jcode_length;j++){
      //cout<<" j="<
      if(Arcode[ch]->code[0][j]){//if true
    //cout<<1;
    //cout<<" "<<1<<" ";
    A[j+seek]=true;
      }
      else{
    //cout<<0;
    //cout<<" 0 ";;
    A[j+seek]=false;
      }
    }
    seek+=j;
    //cout<<<"seek is:"<<
    //cout<
    i++;
}
i=0;
while(i<=size/8+1){
    ch=(char)A.a[i];
    wfile.seekp(i,ios::beg);
    wfile.write(&ch,1);
    i++;
}
//wfile.write((char*)&A.a,size/8+1);                    ///point
//cout<<"in file:"<<
wfile.close();
rfile.close();
//cout<<"file closed"<
}
void inorderWalk(code_ch *root){
if(root->left!=NULL)
    inorderWalk(root->left);
cout<ch<<" "<number<
if(root->right!=NULL)
    inorderWalk(root->right);
}
void decoding(code_ch *root,int size){//very simple only read from file and if 0 go to left else go to right
//cout<<"coming into decodig"<
//cout<<"--------------------------------------------------size="<<
code_ch *p=root;
fstream rfile,wfile;
rfile.open("encoded.txt",ios::in);
wfile.open("decoded.txt",ios::out);
//cout<<"after openning the files"<
if(!rfile || !wfile)
     throw runtime_error("cannot open the files...\n");
rfile.seekg(0,ios::end);
int end=rfile.tellg();
rfile.seekg(0,ios::beg);
int i=0,j=0;
char ch;
//cout<<"before declaring A"<
//cout<<"--------------------------------------------------size="<<
bitArray A(size);
//cout<<"--------------------------------------------------size="<<
//cout<<5<

//cout<<"root hehhhhhhhhhhhhhere :"<left->number<
//cout<<"--------------------------------------------------size="<<
i=0;
while(i<=size/8+1){
    rfile.seekg(i,ios::beg);
    rfile.read(&ch,1);
    A.a[i]=(unsigned char)ch;
    i++;
}
rfile.seekg(0,ios::beg);
i=0;
while(i<=size){//while(!file.eof()){
//cout<<"in while"<
    if(A[i]){//if it is true go to right
      //cout<<1;
      if(p->right!=NULL)
    p=p->right;
      else {
    wfile.seekp(j,ios::beg);
    wfile.write((char*)&p->ch,1);
    j++;
    p=root;
    continue;///point
      }
    }
    else{

      //cout<<"--------------before else"<
      //cout<<0;
      if(p->left!=NULL){
    //cout<<"in if of else"<
    p=p->left;
      }
      else {
    //cout<<"in else of it"<
    wfile.seekp(j,ios::beg);
    wfile.write((char*)&p->ch,1);
    j++;
    p=root;
    continue;///point
      }

    }
    i++;
}
// cout<<9<
rfile.close();
wfile.close();
//cout<<10<
}
int main(int argc,char **argv){
if(argc==1){
    cout<<"you haven't entered a file! exiting..."<
    return 1;
}
listw;
int *a=new int [200];//for saving the number of each ch
for(int i=0;i<200;i++)
    a[i]=0;
bool is_read=read_file(argv[1],a,&w);
if(!is_read)
    throw runtime_error("cannot open the file you entered...\n");
//cout<<"and the heap"<<
PQueue p(w);
code_ch *root=huffman(p,w.size());
find_code(root);
int size=get_size(root);
//cout<<"this is size in the main:"<<
code_ch ** Arcode=inar(root);//,int tree_size);
char *coded;
//cout<<"first puase:"<
//cin>>paus;
int paus1;
encodeing(root,argv[1],Arcode,size);//encodeing(root,argv[1],Arcode,size);
//cout<<" puase2:"<
//cin>>paus;
decoding(root,size);
cout<<"tamam"<
//cout<
return 0;
}

method call chaining cpp

2010-10-23T02:10:00.000-07:00

//in the name of Allah
#include
using namespace std;
class A{
public:
    int a,b,c;
    A(){
    }
//     A seta(){a=1;cout<<"in a"<
//     A setb(){b=2;cout<<"in b"<
// //     A setc(){c=3;cout<<"in c"<
//     A setc(){c=3;cout<<"in c"<
//
    A *seta(){a=1;cout<<"in a"<
    A *setb(){b=2;cout<<"in b"<
//     A setc(){c=3;cout<<"in c"<
    A *setc(){c=3;cout<<"in c"<
    void seta1(){cout<<"in a"<
    void setb1(){cout<<"in b1"<
    void setc1(){cout<<"in c1"<

    int seta2(){cout<<"in a2"<
    int setb2(){cout<<"in b2"<
//     int setc2(){cout<<"in c2"<

    A seta3(){a=0;cout<<"in a3"<
    A setb3(){b=-1;cout<<"in b3"<
    A setc3(){c=-2;cout<<"in c3"<

};
int main(){
A a;
a.seta()->setb()->setc();//a.seta().setb().setc();
cout<<"=======================checkin        zzg===================="<
cout<<"a="<<<"    b="<<<"     c= "<<
//a.seta().setb1().setc1();//invalid use of void
//a.seta2().setb2().setc2();//:24: error: request for member ‘setb2’ in ‘a.A::seta2()’, which is of non-class type ‘int
    a.seta3().setb3().setc3();
cout<<"=======================checking3===================="<
cout<<"a="<<<"    b="<<<"     c= "<<

}

// //in the name of Allah
// #include
// using namespace std;
// class A{
//   public:
//     int a,b,c;
//     A(){
//     }
// //     A seta(){a=1;cout<<"in a"<
// //     A setb(){b=2;cout<<"in b"<
// // //     A setc(){c=3;cout<<"in c"<
// //     A setc(){c=3;cout<<"in c"<
// //
//     A *seta(){a=1;cout<<"in a"<
//     A *setb(){b=2;cout<<"in b"<
// //     A setc(){c=3;cout<<"in c"<
//     A *setc(){c=3;cout<<"in c"<
//     void seta(){cout<<"in a"<
//     void setb1(){cout<<"in b1"<
//     void setc1(){cout<<"in c1"<
//
//     int seta2(){cout<<"in a2"<
//     int setb2(){cout<<"in b2"<
//     int setc2(){cout<<"in c2"<
//
// };
// int main(){
//   A a;
//   a.seta()->setb()->setc();//a.seta().setb().setc();
//   cout<<"=======================checking===================="<
//   cout<<"a="<<<"    b="<<<"     c= "<<
//   //a.seta().setb1().setc1();//invalid use of void
//   //a.seta2().setb2().setc2();//:24: error: request for member ‘setb2’ in ‘a.A::seta2()’, which is of non-class type ‘int
//
// }

c sudoku program exe

2010-05-28T09:51:00.000-07:00

c sudoku program exe

#include
#include
#include
using namespace std;

const int M = 9 + 10;
int a[M][M], t[M];

void err (char c, int n) {
   string s = "soooooooooooooooti";
   if (c == 'c')
       s = "soton ";
   if (c == 'r')
       s = "satr ";
   if (c == 's')
       s = "morabaa ";
   cout << s << n << " eshkal dare\n";
   exit (0);
}

int main () {
   for (int i=1; i<10; i++)
       for (int j=1; j<10; j++)
           cin >> a[i][j];
   //satr
   for (int i=1; i<10; i++) {
       memset (t, 0, sizeof(t));
       for (int j=1; j<10; j++){
           if (t[ a[i][j] ] == 1)
               err ('r', i);
           t[ a[i][j] ] = 1;
       }
   }

   //soton
   for (int j=1; j<10; j++) {
       memset (t, 0, sizeof (t));
       for (int i=1; i<10; i++) {
           if (t[ a[i][j] ] == 1)
               err ('c', j);
           t[ a[i][j] ] = 1;
       }
   }
   char c;
   cout << "mikhay morabaha ro ham bebinam? (y/n)\n";
   cin >> c;
   if (c == 'y') {
       for (int x=1; x<10; x += 3)
           for (int y=1; y<10; y +=3) {
               memset (t, 0, sizeof (t));
               for (int i=x; i < x+3; i++)
                   for (int j=y; j < y+3; j++) {
                       if (t[ a[i][j] ] == 1)
                           err('s', x*10 + y);
                       t[ a[i][j] ] = 1;
                   }
           }
   }
   return 0;
}

Algorithm

#include
#include
#include
using namespace std;

const int M = 9 + 10;
int a[M][M], t[M];

void err (char c, int n) {
    string s = "soooooooooooooooti";
    if (c == 'c')
        s = "soton ";
    if (c == 'r')
        s = "satr ";
    if (c == 's')
        s = "morabaa ";
    cout << s << n << " eshkal dare\n";
    exit (0);
}

int main () {
    for (int i=1; i<10; i++)
        for (int j=1; j<10; j++)
            cin >> a[i][j];
    //satr
    for (int i=1; i<10; i++) {
        memset (t, 0, sizeof(t));
        for (int j=1; j<10; j++){
            if (t[ a[i][j] ] == 1)
                err ('r', i);
            t[ a[i][j] ] = 1;
        }
    }

    //soton
    for (int j=1; j<10; j++) {
        memset (t, 0, sizeof (t));
        for (int i=1; i<10; i++) {
            if (t[ a[i][j] ] == 1)
                err ('c', j);
            t[ a[i][j] ] = 1;
        }
    }
    char c;
    cout << "mikhay morabaha ro ham bebinam? (y/n)\n";
    cin >> c;
    if (c == 'y') {
        for (int x=1; x<10; x += 3)
            for (int y=1; y<10; y +=3) {
                memset (t, 0, sizeof (t));
                for (int i=x; i < x+3; i++)
                    for (int j=y; j < y+3; j++) {
                        if (t[ a[i][j] ] == 1)
                            err('s', x*10 + y);
                        t[ a[i][j] ] = 1;
                    }
            }
    }
    return 0;

}

#include
#include
#include
using namespace std;

const int M = 9 + 10;
int a[M][M], t[M];

void err (char c, int n) {
   string s = "soooooooooooooooti";
   if (c == 'c')
       s = "soton ";
   if (c == 'r')
       s = "satr ";
   if (c == 's')
       s = "morabaa ";
   cout << s << n << " eshkal dare\n";
   exit (0);
}

int main () {
   for (int i=1; i<10; i++)
       for (int j=1; j<10; j++)
           cin >> a[i][j];
   //satr
   for (int i=1; i<10; i++) {
       memset (t, 0, sizeof(t));
       for (int j=1; j<10; j++){
           if (t[ a[i][j] ] == 1)
               err ('r', i);
           t[ a[i][j] ] = 1;
       }
   }

   //soton
   for (int j=1; j<10; j++) {
       memset (t, 0, sizeof (t));
       for (int i=1; i<10; i++) {
           if (t[ a[i][j] ] == 1)
               err ('c', j);
           t[ a[i][j] ] = 1;
       }
   }
   char c;
   cout << "mikhay morabaha ro ham bebinam? (y/n)\n";
   cin >> c;
   if (c == 'y') {
       for (int x=1; x<10; x += 3)
           for (int y=1; y<10; y +=3) {
               memset (t, 0, sizeof (t));
               for (int i=x; i < x+3; i++)
                   for (int j=y; j < y+3; j++) {
                       if (t[ a[i][j] ] == 1)
                           err('s', x*10 + y);
                       t[ a[i][j] ] = 1;
                   }
           }
   }
   return 0;
}

To download   the exe file code

Click Downloads

IF you have comments or ideas leave them below

CPU Cooling methods

2009-09-17T12:04:00.000-07:00

CPU cooling methods

The massive activity that goes on inside your PC generates extensive heat., sometimes known to reach 185 degrees F. The reasons for your PC heating up are
• Insulation caused by dust – dust filters can be provided and other persistent dust can be vacuumed out
• Scant or erratic airflow – can be modulated for even spread

Optimally the CPU has to be maintained around a temperature of 70 degrees F. A CPU coolant does just that. In its absence, overheat leads into myriad problems culminating in frequent crash-downs, computer freezing and in curtailing the life-span of the computer.
A passive heat sink is the basic type of CPU cooler. It draws heat away from the processor, which is dissipated over a large metallic (aluminum is commonly used) surface area. When a fan, atop the sink, or below it, exchanges the heat from the chip, for cooler air, this becomes an active heat sink When suitable computer parts are throttled to bring down the heat it is called soft cooling.
The sink is attached to the processor using spring or screw devices or a thermal interface material that adheres the processor to the heat sink. The interface material should be used in computers, for which mounting the sink on screws or springs is not possible. An inappropriate application of thermal compound can lead to inadequate heat transfer.

A good heat sink characteristics

• Fans (about 12 volts power and 2X120 mm dimension) have to be designed optimally to dissipate the right amount of heat
• The right surface area for heat transfer
• Designed for maximum thermal transfer – a factor dependent on the material of the sink and the thickness of the fins (Thick fins – good conductivity, fine fins – better dissipation)
• A combination of optimal area and the fins spaced at the right distance should be sought
• Flat surface at the heat source ensures an even, thin application of interface material, for better cooling
• At the surface of contact the level of adherence must be high, to provide good pressure

The performance of the sink is measured in C/W or K/W (thermal resistance), not in the regular temperature scales like Centigrade or Kelvin, as difference in temperature is the required parameter

For instance of a load of 30W is applied and the temperature rises by 15 degrees C then the performance is rated as
15C/30 W = ½ C/W.
However you cannot use this parameter as one of the criteria for purchase of your computer, because the thermal resistance displayed on the product may be inaccurate or exaggerated, as a marketing strategy.

Liquid cooling
This consists of a combination of
• a system of pipes with liquid, that run inside the cooling system to extract heat from the processor and cool it
• a pump for liquid circulation. a cooling block to wring out the heat from the microprocessor
• a optional radiator with a condenser coil to decrease the temperature of water and send out the hear from the CPU onto the cooling arrangement
PCs with this set up do not need fans whereas a Cray 2 will need radiators to complete the set up.
The liquid must
• exhibit a lower level of thermal conductivity
• have a certain extent of dielectric nature
Generally the liquids used are motor oils, various other oils including cooking oil, and Flour inert, (a special cooling oil manufactured by the 3M Company)

Water is used significantly in computers set for overclocking. Liquid nitrogen (or sometimes dry ice) cooling can generate a high efficiency rate in computer working. This uses water as the medium to cool/condense… However nitrogen is used only in highly overclocked situations as it needs to be refilled. Besides, the system may succumb to the temperature deviations created, created within the cooling system.

Be wary of
• Heated liquid- This problem is taken care of by fans, usually low-noise ones, to cool the liquid which are set up outside the portable computer’s case.
• Evaporation –When the overheated liquid eventually evaporates, resort to sealing the medium within the computer or refill the liquid whenever needed
• Seepage- For leak problems, disconnect the computer from the power source, mop the seepage with an absorbing cleaning material , identify the source of leakage and replace it… Avoid skin contact with coolant, as it may cause irritation.

The other cooling options and methods are
• Peltier cooling creates a temperature difference using Bismuth-telleride thermocouples, stacked in hundreds on the principle of Seebeck effect
• HVAC systems in large Data Centers
• Phase Change cooling in PCs, situated under it with pipes reaching the processors
• H2Ceramic Cooling –uses sensors to detect overheat-combining with Peltier and liquid cooling

Heat Sensors To help manage CPU cooling the motherboard can be equipped with a smart sensor to indicate voltage, temperature of the CPU and the fans. Smart sensors are now being equipped with features that can, on excess heating
• set off an audio alarm
• flash a warning message on your screen
• shut down the system automatically.

For more related information you can reach to
http://www.heatsink-guide.com/
http://en.wikipedia.org/wiki/CPU_cooling#Spot_Cooling
http://www.pantherproducts.co.uk/Articles/CPU/CPU%20Cooling.shtml
* * * * * * * * * *

A+ Test
1. Should the fans
a)blow air towards the heat sink
b)draws air away from it
c)change to and fro in direction
to cool effectively

2. The fins on the should be
a) Fine and closely spaced
b) Thicker and closely spaced
c) Fine and widely spaced
d) Thick and widely spaced
3. Cooking oil is a good option as a cooling liquid because
a) It is an electrical Insulator
b) It can dispel heat
c) Both the above
d) Neither of the above because It has a very low melting point
4. Cables for airflow in a cooling unit must be
a) Flat ribbon cables to synchronize with the storage drive and holding the conductive wires together tightly, to reduce surface area
b) Rounded cables holding the conductive wires together tightly to reduce surface area
c) Neither of the above
5. Consider these statements regarding water as the cooling liquid for your CPU
A. Freezing point of water used in coolants can be reduced with additives.
B. Color is added to monitor flow
C. Anti corrosive/antimicrobial additives increases its efficiencyas a cooling medium
a) A,B and C are essential
b) A is optional
c) B is Optional
d) C is optional
e) All additives are optional . water works fine without any of them

NUMBER PATTERNS AND SEQUENCES TUTORIAL

2009-09-12T13:08:00.000-07:00

NUMBER PATTERNS AND SEQUENCES TUTORIAL

Subject : Mathematics Form 1

and To get the document in MS word file

=
View

To Get more DOWNLOAD

IN WORD DOCS

or see it on gooogle doc

Learning Objective : Number Patterns and Number Sequences Learning Outcomes : 1. Identify patterns of number sequences. 2. Extend, complete and construct number sequences. 3. Recognize odd and even numbers and explore their general properties. 4. Identify prime numbers. 5. Understand factors and prime factors. 6. Find the common factors and highest common factors (HCF). 7. Understand multiples. 8. Find the common multiples and lowest common multiples (LCM

2.1 Number Patterns & Number Sequences - A list of numbers that follow a certain pattern is called number sequence. - In a number sequence, we can see how the number pattern is form.

To see the lesson in MS word form
DOWNLOAD hERE

1. 3, 7, 11, 15, 19 Pattern : begin with 3 and add 4 to the number before it (+4) 2. 305, 300, 295, 290, 285 Pattern : begin with 305 then minus / subtract 5 from the number before it (- 5) 3. 2, 6, 18, 54, 162 Pattern : begin with 2, then multiply each number by 3 (x3) 4. 64, 32, 16, 8, 4 Pattern : begin with 64 then divide each number by 2 (÷ 2)

Complete the missing number in the n

List the number sequences for these number sequences 1. 3, 6, 9 , 12, 15 , 18 Pattern : + 3 2. 64, 56, 48, 40, 32, 29 Pattern : - 8 3. 7, 21, 63, 189, 567, 1761 Pattern : x 3 4. 800, 400, 200, 100, 50, 25 Pattern : ÷ 2

umber patterns

1. List down the whole numbers between 30 to 37 31, 32, 33, 34, 35, 36 2. Add 5 to whole numbers from 3 to 28 3, 8, 13, 18, 23, 28 3. Subtract 3 from whole number from 13 to 1 13, 10, 7, 4, 1 4. Multiply 4 to whole numbers from 2 to 128 2, 8, 32, 128

2.2 Even Numbers & Odd Numbers

- Even numbers are whole numbers that can be divided by 2 exactly (no remainder). Example: 2, 4, 6, 8, 10, … - Odd numbers are whole numbers that cannot be divided by 2 exactly (has remainder). Example: 1, 3, 5, 7, 9, 11, … - ‘0’ is neither an add number nor an even number Determine whether these numbers are even numbers or odd numbers 1. 214 214 ÷ 2 = 107 therefore 214 is an even number 2. 735 735 ÷ 2 = 367 remainder 1 therefore 735 is an odd number 3. 2 579 2 579 ÷ 2 = 1 289 remainder 1 therefore 2 579 is an odd number 4. 5 550 5 550 ÷ 2 = 2 775

2.3 Prime Numbers

- Prime number is a whole number that can only be divided by itself and number 1 - The number 1 is not a prime number because it can only be divided by itself - All the prime numbers are odd numbers except for 2 - Example : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, … Determine whether this number is a prime number or not 1. 31 31 ÷ 1 = 31 31÷ 31 = 1 31 can only be divided by 1 and itself. 31 is a prime number 2. 65 65 ÷ 1 = 65 65 ÷ 65 = 1 65 ÷ 5 = 13 65 can be divided by 1, itself and also 5. 65 is not a prime number 3. 71 71 ÷ 1 = 71 71÷ 71 = 1 71 can only be divided by 1 and itself. 71 is a prime number 4. 93 93 ÷ 1 = 93 93 ÷ 93 = 1 93 ÷ 3 = 31 93 can be divided by 1, itself and 3. 93 is not a prime number

Sieve Of Erastosthenes - A method of finding prime numbers between 1 to 100 (25 numbers) Step 1: list down all whole numbers between 1 to 100

Whole numbers photo

Step 2: Cross out 1, because 1 is not a prime number. Step 3: Circle 2 and cross out all numbers that can divided by 2. Step 4: Circle 3 and cross out all numbers that can divided by 3. Step 5: Circle 5 and cross out all numbers that can divided by 5. Step 6: Circle 7 and cross out all numbers that can divided by 7. Step 7: Circle all remaining numbers and list down. The remaining numbers are the prime numbers between 1 to 100. Answer: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

http://www.4shared.com/file/125589473/1f7766d0/math_tests__School_exams_e.html

2.4 Factors

- A factor of a given number is the number that can divide the given number exactly without any remainder. - The number 1 is a factor of all numbers. - Every number is a factor of itself - A whole number may have more than 2 factors. List all the factors of these numbers 1. 6 1 x 6 factors of 6 = 1, 2, 3, 6 2 x 3 2. 18 1 x 18 factors of 6 = 1, 2, 3, 6, 9, 18 2 x 9 3 x 6 3. 45 1 x 45 factors of 6 = 1, 3, 5, 9, 15, 45 3 x 15 5 x 9 4. 88 1 x 88 factors of 6 = 1, 2, 4, 8, 11, 22, 44, 88 2 x 44 4 x 22 8 x 11 Determine whether 1. 9 is a factor of 54 54 ÷ 9 = 6 (exact division, no remainder) Therefore 9 is a factor of 54 2. 7 is a factor of 48 48 ÷ 7 = 6 remainder 6 (not exact division) Therefore 7 is not a factor of 48

2.5 Prime Factors

- Prime factors of a given number are factors which are also prime numbers. - Example: Factors of 6: 1, 2, 3, 6 Prime numbers: 2 and 3 Prime factors of 6: 2 and 3 List all the prime factors of these numbers. 1. 24 Method 1 : List the factors Factors of 24 : 1, 2, 3, 4, 6, 8, 12, 24 Prime Factors : 2 and 3

Method 2 : Continuous Division

https://docs.google.com/exercises

Determine whether

1. 2 is a prime factor of 18 18 ÷ 2 = 9 (exact division, no remainder) Therefore 2 is a prime factor of 18 2. 7 is a factor of 46 46 ÷ 7 = 6 remainder 4 (not exact division) Therefore 7 is not a prime factor of 46 3. 4 is a factor of 200 200 ÷ 4 = 50 (exact division, no remainder) 4 is a factor of 200 but 4 is not a prime number Therefore 4 is not a prime factor of 200

.6 Common Factors & Highest Common Factors (HCF)

Common Factors - Common factor is a number that is a factor of two or more numbers. - 1 is a common factor of all numbers. Find all the common factors of these numbers. 1. 8 and 12 Factors of 8 : 1, 2, 4, 8 Factors of 12 : 1, 2, 3, 4, 6, 12 Common factors of 8 and 12 : 1, 2, 4 2. 6, 12 and 18 Factors of 6 : 1, 2, 3, 6 Factors of 12 : 1, 2, 3, 4, 6, 12 Factors of 18 : 1, 2, 3, 6, 9, 18 Common factors of 6, 12 and 18: 1, 2, 3, 6 3. 27, 36 and 81 Factors of 27 : 1, 3, 9, 27 Factors of 36 : 1, 2, 3, 4, 6, 9, 12, 18, 36 Factors of 81 : 1, 3, 9, 27, 81 4. Common factors of 27, 36 and 81: 1, 3, 9 Determine whether 1. 6 is a common factor of 12, 18 and 24 12 ÷ 6 = 2 (exact division, no remainder) 18 ÷ 6 = 3 (exact division, no remainder) 24 ÷ 6 = 4 (exact division, no remainder) Therefore 6 is a common factor of 12, 18 and 24 2. 9 is a common factor of 63 and 120 63 ÷ 9 = 7 (exact division, no remainder) 120 ÷ 9 = 13 remainder 3 (not exact division) Therefore 9 is a not common factor of 63 and 120
Highest Common Factors ( HCF )
- HCF of two or more numbers is the largest common factor of these numbers.

2.7 Multiples

- The multiples of a number is the product of that number with any whole number except

zero.

- Multiples are also a sequence.

List the first five multiples of these numbers.

1. 3

3 x 1 = 3

3 x 2 = 6

3 x 3 = 9

3 x 4 = 12

3 x 5 = 15

The first five multiples of 3 are 3, 6, 9, 12 and 15

2. 9

9 x 1 = 9

9 x 2 = 18

9 x 3 = 27

9 x 4 = 36

9 x 5 = 45

The first five multiples of 3 are 9, 18, 27, and 45

List all the multiples of these numbers.

1. Multiples of 2 between 13 to 27

14, 16, 18, 20, 22, 24, 26

2. Multiples of 5 from 50 to 70

50, 55, 60, 65, 70

Determine whether

1. 48 is a multiple of 4

48 ÷ 4 = 12 (exact division, no remainder)

Therefore 48 is a multiple of 4

2. 26 is a multiple of 3

26 ÷ 3 = 3 remainder 2 (not an exact division, has remainder)

Therefore 26 is not a multiple of 3

2.6 Common Multiples & Lowest Common Multiples (LCM)

Common Multiples
- Common multiple is a number that is a multiple of two or more numbers.

- Example 8 is a common multiple of 2 and 4

Multiple of 2 : 2, 4, 6, 8, 10, 12 ,… ( 8 is multiple of 2 )

Multiple of 4 : 4, 8, 12, 16, 20, … ( 8 is multiple of 4 )

Therefore 8 is a common multiple of 2 and 4

Find the first three common multiples of these numbers.

1. 2 and 3

Multiple of 2 : 2, 4, 6, 8, 10, 12 , 14, 16, 18, 20, …

Multiple of 3 : 3, 6, 9, 12, 15, 18, 21, 24, …

Therefore the first three common multiples of 2 and 3 are 6, 12, and 18

2. 3, 4 and 6

Multiple of 3 : 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36,..

Multiple of 4 : 4, 8, 12, 16, 20, 24, 28, 32, 36, ..

Multiple of 6 : 6, 12, 18, 24, 30, 36, ..

Therefore the first three common multiples of 3, 4 and 6 are 12, 24 and 36

Determine whether

1. 50 is a common multiple of 2 and 5

50 ÷ 2 = 25 (exact division, no remainder)

50 ÷ 5 = 10 (exact division, no remainder)

Therefore 50 is a common multiple of 2 and 5

2. 120 is a common multiple of 3, 4 and 9

120 ÷3 = 40 (exact division, no remainder)

120 ÷4 =30 (exact division, no remainder)

120 ÷ 9 = 13 remainder 3 (not exact division)

Therefore 120 is not a common multiple of 3, 4 and 9

http://www.4shared.com/file/125589473/1f7766d0/math_tests__School_exams_e.html

Lowest Common Multiples ( LCM )
- LCM of two or more numbers is the smallest common multiple of these numbers.

Find the LCM of the followings.

1. 6 and 36

Method 1 : Listing the multiples

Multiple of 6 : 6, 12, 18, 24, 30, 36, .

Multiple of 9 : 9, 18, 27, 36, …

The LCM of 6 and 9 is 18

and a math test https://docs.google.com/math exercises
whole number lesson

https://docs.google.com/leaf?id=1WxPw2LhqeyNK0wrONdEjm2bgsBcPadqKXloY6ZxdiAx2XFfAncqb7TOtrkuh&sort=name&layout=list&num=50

Please leave your comments and notices here below

Mathematics problems sheet

2009-08-30T05:29:00.000-07:00

Mathematics problems sheet

SEKOLAH MENENGAH KEBANGSAAN BANDAR BINTULU PEPERIKSAAN PENGGAL PERTAMA (PP1) / 2008 MATEMATIK TINGKATAN 1 MASA : 1 JAM 15

Name: ……………………………………………. Class: ………………….

This paper consists of 40 questions. Answer all the questions. Each question is followed by four options A, B, C and D. For each question, choose the most appropriate answer and blacken your answer on objective answer sheet. You are not allowed to use calculator for this examination.

DOWNLOADS

1. The number 513 579 is written as 514 000 when it is rounded off to the nearest
A. ten
B. hundred
C. thousand
D. ten thousand

2. Which of the following is equal to 4 558?
A. 4 000 + 550 + 8
B. 4 000 + 500 + 8
C. 4 500 + 550 + 8
D. 4 550 + 500 + 8

3. The diagram above shows a group of numbers. Find the difference between the largest and the smallest numbers.
A. 18
B. 90
C. 99
D. 189

4. When the number 508 and 79 are subtracted from 6 013, we get
A. 5 246
B. 5 426
C. 5 546
D. 6 442

5. [ 754 – ( 345 – 167 ) × 3 ] ÷ 11
A. 175
B. 117
C. 108
D. 20'

6. Norman bought 15 boxes of apples. Each box contains 32 apples. If all of the apples are to be distributed equally among 24 children in an orphanage, how many apples will each child get?
A. 10
B. 16
C. 20
D. 30

7. A factory produces 1 560 pairs of shoes per day. In a particular week, 278 pairs were spoilt. If the factory operates every day, how many pairs of shoes produced in that week were not spoilt?
A. 8 974
B. 10 642
C. 10 880
D. 12 866

8. Mrs. Ramani has 80 pencils, 28 erasers and 60 rulers. She wants to prepare some gift packs; each pack containing 5 pencils, 2 erasers and 4 rulers. Find the maximum number of gift packs that she can prepare.
A. 14
B. 15
C. 16
D. 17

9. In the number sequence, 2, 5, 11, x, 32, 47, …, the value of x is
A. 15
B. 18
C. 20
D. 24

10. 11, 13, m, 19, n are prime numbers arranged in ascending order. The value of m + n is
A. 36
B. 38
C. 40
D. 46

11. x is a common multiple of 3 and 39. Which of the following is not x ?
A. 78
B. 117
C. 156
D. 189

12. Find all the prime factors of 102.
A. 2, 3 and 7
B. 2, 3 and 11
C. 2, 3 and 17
D. 3, 7 and 11

13. The lowest common multiple (LCM) of 4, 6 and x is 36. The possible value of x is
A. 18
B. 24
C. 48
D. 56
14. The lowest common multiple (LCM) of which number set below is not 12 ?
A. 4, 6, 8
B. 3, 4, 6
C. 2, 3, 4
D. 3, 4

15. Find the highest common factor (HCF) of 16, 24 and 40.
A. 6
B. 8
C. 12
D. 16
16. Given that ( m–3 ) is the highest common factor (HCF) of 32 and 40, find the value of m.
A. 5
B. 8
C. 11
D. 163

22. Given that 4/7 of the students in a class are girls, find the total number of students if 15 of them are boys.
A. 19
B. 20
C. 35
D. 40
23. Farah is given RM320 by her father. 1/8 and of the money are used to buy pens and books respectively. How much money has Farah left?
A. RM 40
B. RM 128
C. RM 152
D. RM 160

25. Round off 8.0392 to 2 decimal places.
A. 8.03
B. 8.04
C. 8.05
D. 8.10

26. The diagram above shows number cards. On which of the above number cards does digit 3 have the same value?
A. P and Q
B. P and S
C. Q and R
D. Q and S
27. The value of 20.05 + 2.005 rounded off to 1 decimal place is
A. 22.0
B. 22.06
C. 22.1
D. 22.10

28. Which of the following decimals are in descending order?
A. 0.007, 0.017, 0.7, 1.07, 1.70
B. 1.70, 1.07, 0.7, 0.017, 0.007
C. 1.07, 1.70, 0.7, 0.007, 0.017
D. 1.70, 0.7, 0.017, 0.007, 1.07

29. Which of the following is not true?
A. 3.092 × 100 = 309.2
B. 0.2 × 0.4 = 0.8
C. 7.59 × 10 = 75.9
D. 0.5 × 20 = 10

30. If 4.5 kg of oranges cost RM 12.60, calculate the price of 3.2 kg of oranges.
A. RM 6.00
B. RM 8.96
C. RM 13.90
D. RM 20.30
31. The product of three numbers is 11.88 . If two of the numbers are 3.3 and 2, what is the third number?
A. 1.8
B. 2.4
C. 2.6
D. 3.1

32. Aida, Siew May and Betty donated RM164.45 to a charity fund. If Siew May and Betty donated the same amount and Aida donated RM5.75 more than Betty, what was the amount donated by Aida?
A. RM 52.90
B. RM 58.65
C. RM 79.35
D. RM 85.10

33. Convert into percentage.
A. 45%
B. 60%
C. 75%
D. 80%
34. 38% of 250 is
A. 85
B. 93
C. 95
D. 155

35. If p × 8 = 0.12, what is the value of p?
A. 1.5%
B. 2.0%
C. 2.5%
D. 3.0%

36. 90 out of 225 in percentage is
A. 15%
B. 20%
C. 25%
D. 40%
37. Mamat’s monthly salary is RM 2400. He spends RM 1824 each month and saves the balance of the salary in the bank. What percentage of his salary is saved?
A. 24%
B. 25%
C. 26%
D. 27%

38. A total of 200 pupils visited a public library on a certain Friday. 40% of them were girls. The number of boys who visited the library increased by 50% the next day. Calculate the total number of boys who visited the library on the two days, Friday and Saturday?
A. 60
B. 80
C. 180
D. 300
39. The price of a school bag is RM 95. Find the price of the school bag after a 20% discount.
A. RM 19
B. RM 66
C. RM 76
D. RM 114

40. Table above shows the number of foreign workers employed by a factory in 3 years. Which of the following statements is false?
A. 80% of the workers in year 2001 are
foreign workers.
B. 28 % of the workers in year 2002 and
2003 are not foreign workers.
C. The percentage of foreign workers in years 2002 and 2003 is the same.
D. Year 2001 has the lowest percentage of
foreign workers.

Answer scheme sheet

Math Test

2009-08-22T02:01:00.000-07:00

In Math test

Download Now

SECOND PROGRESSIVE TEST (UK 2) / 2009
MATHEMATICS
FORM 1
TIME : 1 HOUR
PREPARED BY : CIK HJH NOR HAZRINA HJ HADZELAN

AME : _________________________________ FORM : ____________

SECTION A ( 40 Marks )
This section contains 20 objective questions. Answer all questions. Each question is followed by four answer choices A, B, C and D. Choose one answer only. Write your answers in space provided. Calculator is not allowed.

1. The place value of digit 1 in the number 513 029 is
A. Hundreds
B. Thousands
C. Ten thousands
D. Hundred thousands

2. Among the following numbers, which becomes 632 000 when rounded off to the nearest thousand?
A. 631 099
B. 631 201
C. 631 487
D. 631 500

3. The diagram above shows a group of numbers. Find he difference between the largest and the smallest number.
A. 199
B. 450
C. 504
D. 549

4. The remainder of 7 739 18
A. 17
B. 19
C. 349
D. 429

5 7 6
8 9
5 1 8 4
+
____________
5 1 2 6 4

5. The calculation above shows the multiplication of 576 and 89. What is the number in the box?
A. 4 068
B. 4 568
C. 4 608
D. 4 768

6. Saiful bought 24 boxes of pens and each box contains 96 pens. He sold the pens at RM 3 each. Calculate the amount of money he got from selling all the pens.
A. RM 2 304
B. RM 5 550
C. RM 6 912
D. RM 7 300

7. 48 + 15 60 12 =
A. 63 5
B. 48 + 75
C. 25 + 15 5
D. 4 + 15 60

8. 192 3 ( 6 + 2 )
A. 8
B. 48
C. 386
D. 512

9. Ling has 49 marbles, which are 19 less than Muthu but 13 more than Nazrul. To start a game, all marbles from three players are collected and divided equally among them. The number of marbles each player will have is
A. 39
B. 43
C. 51
D. 57

10. A school buys 310 boxes of pencils. There are 18 pencils in each box. After giving every student 4 pencils each, there are 180 pencils left. The number of students in the school is
A. 1 350
B. 1 440
C. 5 440
D. 5 760

11. 109, 89, 76, x , 50, ...
The value of x in the number sequence
above is
A. 60
B. 63
C. 65
D. 79

12. 1, 2, 5, p, 17, 26, q, ...
The list of numbers above is a number sequence. The value of p + q is
A. 35
B. 42
C. 44
D. 47

13. m, 37, 41, 43, n, ...
The list of numbers above is a sequence of prime numbers in ascending order. The value of n - m is
A. 10
B. 12
C. 14
D. 16

14. Given that 8 and 9 are the factors of Y. What is the possible value of Y?
A. 63
B. 96
C. 126
D. 144

15. Given that 2 and 3 are the prime factors of 132. The other prime factor of 132 is
A. 11
B. 9
C. 7
D. 5

16. Which of the following numbers is the lowest common multiple (LCM) of 6, 8 and 12?
A. 12
B. 24
C. 48
D. 72

17. 112 is the common multiple of 8 and x. Among the following, which is not a value of x ?
A. 14
B. 18
C. 28
D. 56

18. The total number of common factors of 12, 18 and 48 is
A. 1
B. 3
C. 4
D. 6

19. Which of the following pairs of numbers has 6 as the highest common factor (HCF)?
A. 12 and 36
B. 18 and 24
C. 18 and 36
D. 42 and 84

20. Given that ( x – 7 ) is the highest common factor of 24 and 32. Find the value of x.
A. 4
B. 8
C. 11
D. 15

SECTION B ( 60 Marks )

This section contains 10 subjective questions. Answer all questions. Write your answers in space provided. All the workings must be shown. Calculator is not allowed.

1. a) Write 2 950 611 in words.

(2 marks)

1. b) State the place value and the value of digit
9 in 390 451

Place Value:

Value of digit :

(4 marks)

2. Round off 796 145 to the nearest
a) thousands

(2 marks)
b) ten thousands

(2 marks)
c) hundred thousands

. Calculate the followings

a) 26 705 + 57 + 6 998 =

(3 marks)
b) 96 24 =

(3 marks)

(2 marks)

4. Calculate the followings

a) 54 + 1 575 25 =

(3 marks)
b) 110 – 3 ( 16 + 24 4 ) =

(3 marks)

The method above is used to find the Lowest
Common Multiples (LCM) of 54 and n .
Find the value of n  m

6. Find the sum of Highest Common Factor (HCF)
of 12 and 18 and Highest Common Factor (HCF)
of 12 and 24.

7. a) Find the value of
Give your answer correct to 2 decimal places.

7. b) + 13 ´ 1.25

8. a) Write 32 % as a fraction in its lowest term.

9. a) Find the number if 4.5 % of a number is 27.
9. b) Find the final value of 900 increased by 5%

10. Susan has 30 metres of fabric. She used 18.4
metres of the fabric and sells the remainder to
Mimi for RM 3.50 per metre. How much does
Mimi has to pay to Susan?

11. Lee bought a bicycle for RM 400. Later he

sold the bicycle to Rama for a profit of RM 70.

Find the percentage of profit.

12. a) Arrange the following in decreasing order.

-7 , 4 , -11 , 16 , -2

12. b) Figure below shows a number sequence.
Determine the value of m + n .

20 , m , -10 , -5 , 0 , n , 10

14. The temperature of a town in the morning is
3C. During the afternoon the temperature
increases 11C and later at night it drops 25C.
What is the temperature during the night?

15. a) Mr. f drives a distance of k km from
City p to City q in just 30 minutes.
State the unknown.

15. b) Pick out the like terms in the following.

10 p , -14 r , 9.3 y , r , -8 u

16. a) 10 h + 2 - 16 - 6 h

17. Julie buys a story book which costs RM y. She
pays with a RM 50 note. Write an algebraic
expression to shows the balance of money she
will receive

19. a) A journey from Bintulu to Kuching takes
5 hours. A bus departs from Bintulu at
10.30 a.m. State the time in the 24-hour
system that the bus arrives at Kuching.

19 b) The weight of 1 packet of sugar is 495 g.
If Komala buys 8 packets of sugar, what is the
total weight of sugar, in kg , Komala bought?

20. Figure below shows the various routes a traveller can takes to go from town A to town D.
Calculate the difference between the longest route and the shortest route to go from town A to town D.

1. a) Write 2 750 438 in words. b) State the place value and the value of
digit 9 in 597 143

(2 marks) (4 marks)

2. Round off 818 795 to the nearest
a) ten. b) thousand. c) hundred thousand.

(2 marks) (2 marks) (2 marks)
3. a) Find the sum of 207, 4 659 and 13 022. b) Find the difference between 97 501 and 899.

(3 marks) (3 marks)
4. Table below shows the number of visitors who visited Zoo Negara from March to June
in a particular year.
Month March April May June
Number of visitors 4 079 3 788 2 413 3 990

a) Calculate the total number of visitors from b) Round off the total number of visitors to the
March to June. nearest hundred.

(4 marks) (2 marks)
5. Calculate the followings.
a) 592 8 b) 405 16 c) 11 172 12

(2 marks) (3 marks) (3 marks)
6. If 4 tailors can each sew 8 shirts in a day, how many shirts can they sew altogether in 3 weeks?

(4 marks)
7. Calculate the followings.
a) 45 + 135 5 - 16 b) 114 3 + 7 12 c) 95 – 2 ( 12 + 42 7 )

(3 marks) (3 marks) (3 marks)
8. Siew Lan had RM 107. She bought a story book for RM 14 . She also bought eight boxes of
colour pencils which is cost RM 6 each. Find the balance of her money.

(3 marks)
9. a) Complete the following number sequence.

5, 11, 17, ______, ______, ______, 41
(3 marks)

b) In the number sequence below, find the value of x + y

1, 2, 4, x , 16, y

(3 marks)
10. a) List out all the odd numbers between 44 and 52.

(2 marks)
b) Find the sum of all the even numbers that are less than 9.

(2 marks)
a) List out the first two prime numbers that are greater than 90.

(2 marks)

SECTION B ( 60 Marks )
This section contains 10 subjective questions. Answer all questions. Write your answers in space provided. All the workings must be shown. Calculator is not allowed.

1. a) Write 2 750 438 in words. b) State the place value and the value of
digit 9 in 597 143
two million seven hundred and fifty thousand
four hundred and thirty-eight place value : ten thousands
value of digit : 90 000

(2 marks) (4 marks)
2. Round off 818 795 to the nearest
a) ten. b) thousand. c) hundred thousand.

818 800 819 000 800 000

(2 marks) (2 marks) (2 marks)
3. a) Find the sum of 207, 4 659 and 13 022. b) Find the difference between 97 501 and 899.

207 + 4 659 + 13022 97 501 - 899
= 17 888 = 96 602

(3 marks) (3 marks)
4. Table below shows the number of visitors who visited Zoo Negara from March to June
in a particular year.
Month March April May June
Number of visitors 4 079 3 788 2 413 3 990

a) Calculate the total number of visitors from b) Round off the total number of visitors to the
March to June. nearest hundred.

4 079 + 3 788 + 2 413 + 3 990 14 300 visitors
= 14 270 visitors

(4 marks) (2 marks)
5. Calculate the followings.
a) 592 8 b) 405 16 c) 11 172 12

(2 marks) (3 marks) (3 marks)
6. If 4 tailors can each sew 8 shirts in a day, how many shirts can they sew altogether in 3 weeks?

4 8 7 3 = 672 shirts

(4 marks)
7. Calculate the followings.
a) 45 + 135 5 - 16 b) 114 3 + 7 12 c) 95 – 2 ( 12 + 42 7 )

45 + 27 – 16 38 + 7 X 12 95 – 2 ( 12 + 6 )
= 72 – 16 = 38 + 84 = 95 - 2 ( 18 )
= 56 = 122 = 95 - 36
= 59

(3 marks) (3 marks) (3 marks)
8. Siew Lan had RM 107. She bought a story book for RM 14 . She also bought eight boxes of
colour pencils which is cost RM 6 each. Find the balance of her money.

RM 107 – RM 14 – ( RM 6 8 )
= RM 107 – RM 14 – RM 48
= RM 45

(3 marks)
9. a) Complete the following number sequence.

5, 11, 17, __23____, __29____, ___35___, 41 ( pattern : +6 )
(3 marks)

b) In the number sequence below, find the value of x + y

1, 2, 4, x , 16, y ( pattern : 2 )

x = 8, y = 32 x + y = 8 + 32 = 40

(3 marks)
10. a) List out all the odd numbers between 44 and 52.
45, 47, 49, 51
(2 marks)
b) Find the sum of all the even numbers that are less than 9.
2 + 4 + 6 + 8 = 20

(2 marks)
a) List out the first two prime numbers that are greater than 90.
97 and 101

(2 marks)

Information Technology Terms

2009-07-29T00:17:00.000-07:00

In following article we discussed some terms and issues related to Information technology

Information-processing technology

Information-processing technology, or just information technology (IT), is the most common single type of technology within operations and includes any device which collects, manipulates, stores or distributes information. Often organizational and operational issues are the main constraints in applying information technology because managers are unsure how best to use the potential in the technology. The following quotation gives some idea of how fast information technology has changed: The rate of progress in information technology has been so great that if comparable advances had been made in the automotive industry, you could buy a Jaguar that would travel at the speed of sound, go 600 miles on a thimble of gas and cost only $2!

Centralized and decentralized information processing

All computers used for management purposes were, at one time, large and centralized. It was simply the most economical way of buying processing power. Then the cost and power of smaller computers reached the point where it was economical feasible for different parts of the operation to have their own dedicated computer under the direct control of the staff who would use them. This is the distributed processing concept. The obvious problem with such an arrangement was that, in bringing computing power closer to its users, coordinating all the various processing activities became more complex. The answer to the problem was for the distributed computers to exchange information. This eventually led to the concept of the network.

Telecommunications and information technology

Computer-based technologies in business use have always been based on digital principles (converting information into a binary form using 0s and 1s). Telecommunications, mean while were originally based on analogue technology. The digitization of telecommunications transmissions (including digital compression techniques, which allow information to be squeezed into a smaller ‘space’ so that more can be sent using a given amount of transmission capacity), together with the use of high-capacity optical fibre networks, brought new possibilities. The technologies of computing and telecommunications in effect merged. Digital telecommunication lines could carry both voice and non-voice (text, data, etc.) traffic at the same time, so separate sites of the same organization, or separate operations, could lease lines for their exclusive use. Alternatively, separate operations could use one of the public integrated services digital networks (ISDNs).

The internet

Undoubtedly the most significant technology to impact on operations management in the last few years has been the internet. In effect, the internet is a ‘network of networks’. It is used to link computer networks with other computer networks. Its origin lies in the development of LANs in the 1970s and 1980s (and later, wide area networks, WANs). However, because they used different types of computer, LANs usually found it difficult to talk to each other. Nor did WANs use the same language as LANs. The breakthrough came with the development of a technique called ‘packet switching’. This enabled many messages to be sent to different locations at the same time and allowed individual networks to communicate. In practical terms, though, most of us think of the internet as the provider of services such as the ability to browse the World Wide Web.

E-business

The use of internet –based technology, either to support existing business processes or to create entirely new business opportunities, has come to be known as e-business. The most obvious impact has been on those operations and business processes that are concerned with the buying and selling activity (e-commerce). The internet provided a whole new channel for communicating with customers. The advantage of internet selling was that it increased both reach (the number of customers who could be reached and the number of items they could be presented with) and richness (the amount of detail which could be provided concerning both the items on sale and customers’ behaviour in buying them). Traditionally, selling involved a trade-off between reach and richness. The internet effectively overcame this trade-off. However, the internet had equally powerful implications for the ongoing provision of services.

M-business

The major impact of the internet on so many areas of business has been further boosted by developments in mobile technology. M-business is the phrase now frequently used to cover applications that combine broadband internet and mobile technology devices. For example, some financial services offer their customers access to their accounts through personal digital assistance (PDAs) and mobile (cell) phones. But in business, applications are not limited to enhanced customer service. Generally, communications between staff, especially those who spend much of their time away from the operations, such as sales people, can be significantly facilitated. Mobile communications of this type offer the potential for significant cost savings as well as new business opportunities. However, as with all wireless applications, security concerns can prove a problem in some applications.

Management information systems (MIS)

Within the configuration of any information-processing technology, what is important is the way in which information moves, is changed, is manipulated and presented so that it can be used in managing an organization. These systems are management information systems. Operations managers make considerable use of MISs, especially in their planning and control activities. Systems which are concerned with inventory management, the timing and scheduling of activities, demand forecasting, order processing, quality management and many other activities are an integral part of many operations managers’ working lives and are referred to in the planning and control.

Decision support systems (DSSs)

A decision support system is one which provides information with the direct objective of aiding or supporting managerial decision making. It does this by storing relevant information, processing it and presenting it in such a way as to be appropriate to the decision being made. In this way, it supports managers by helping them to understand the nature of decisions and their consequences, but it does not actually make the decision itself. Often DSSs are used for ‘what if’ analyses which explore the (often financial) consequences of changing operations practice.

Expert systems (ESs)

Expert systems take the idea of DSSs one stage further in that they attempt to ‘solve’ problems that would normally be solved by humans. An ES exhibits (within a specified area) a sufficient degree of expertise to mimic human problem solving. The key part of an ES is its ‘inference engine’ which performs the reasoning or formal logic on the rules that have been defined as governing the decision. These rules are called the ‘knowledge base’ of the ES (which is why ESs are also called knowledge-based systems). There have been many attempts to utilize the idea of an ES in operations management. Although authorities agree that ESs will become far more important in the future of operations management, not all applications so far have been totally successful. The problems which have been encountered include the following:

* Most expert systems can treat only narrow problems rather than the more realistic issues of integration and conflict between problem areas of the operation.
* Putting even some of an operations manager’s expertise into a knowledge base is very expensive in terms of time and processing power.
* Like all information-based systems, it is rendered impotent if the data it is working with are wrong or inaccurate.

The elements of job design

Job design involves a number of separate yet related elements which when taken together define the jobs of the people who work in the operation. Whether you are managing an Egyptian quarry, providing adventure holidays, running a software consultancy or a tax advice office, or building cars, there are six key elements of job design that you will need to consider.

What are the environmental conditions of the workplace?

The conditions under which jobs are performed will have a significant impact on people’s effectiveness, comfort and safety. This is called ergonomic environmental design. It is concerned with issues such as noise, heat and light in the workplace.

What technology is available and how will it be used?

The vast majority of operational tasks require the use of technology, even if the technology is not sophisticated. Not only does the technology need to be appropriate and designed well, so does the interface between the people and the hardware. This is called ergonomic workplace design.

What tasks are to be allocated to each person in the operation?

Producing goods and services involves a whole range of different tasks which need to be divided between the people who staff the operation. Different approaches to the division of labour will lead to different task allocations.

What is the best method of performing each job?

Every job should have any approved method of completion and this should be the ‘best’ method. Although there are different ideas of what is ‘best’, it is generally the most efficient method but that fits the task and does not unduly interfere with other tasks. This is usually referred to as work study – one element of scientific management.

How long will it take and how many people will be needed?

The second element of scientific management is work measurement. Work measurement helps us calculate the time required to do a job so that we can then work out how many people we will need.

How do we maintain commitment?

Keeping staff motivated is not easily. There is a danger that in considering the previous questions it may be tempting to see the person as a unit of resource rather than a human being with feeling and emotions. So understanding how we can encourage people and maintain their commitment is the most important of the issues in job design and work organization. This is concerned with the behavioural approaches to job design including empowerment, teamwork and flexible working.

End of Information Technology terms

c samples program

2009-07-14T07:05:00.000-07:00

In this c sample program compile the following codes and see the output

/*
PROG: butter
LANG: C++
*/

download c_heap_sort
//In the name of God
#include
#include
#include
using namespace std;

const int P = 800 + 10, N = 500 + 10, inf = P * N * 225;
int h[P], nh = 0, ver[P], ind[P], t[30], hh;
vector v[P];

inline void my_swap (int x, int y) {
swap (ver[x], ver[y]);
swap (ind[ ver[x] ], ind[ ver[y] ]);
swap (h[x], h[y]);
}

//x == index
void bobble_up (int x) {
if (x != 1 && h[x] < x ="="" mc =" x;"> h[x*2] && x*2<=nh) mc = x*2; if (h[mc] > h[x*2 + 1] && x*2+1 <= nh) mc = x*2+1; if (mc != x) { my_swap (x, mc); bobble_down (mc); } } //x != index && x == value inline void insert_key (int x) { h[++nh] = x; bobble_up (nh); } inline int extract_min () { int r = h[1]; my_swap(1, nh--); bobble_down(1); return r; } //x == index inline void modifi_key (int x, int val) { h[x] = val; bobble_up (x); bobble_down (x); } int main () { int x; while (cin >> x) {
insert_key (x);
t[hh++] = x;
cerr << x << endl;
}
cerr << "hello\n";
sort (t, t+hh);
/*for (int i=1; i
cerr << h[i] << " ";
cerr << endl;*/
for (int i=0; i
cerr << t[i] << " ";
cerr << endl;
while(nh)
cerr << extract_min () << " ";
cerr << endl;
return 0;
}

copy the code above
The end of the c sample program
for more samples see other posts in the blog
please leave messages and comment below

history of cpr and definition essay

2009-07-08T07:36:00.000-07:00

History of cpr and definition

Definition

which is cardiopulmonary resuscitation, is a technique that is used as a life saving skill that helps keep someone who is suffering from cardiac arrest alive long enough to get to the hospital for treatment. This first aid skill had started to become available in the early 18th century and it was turned into an effective life saving method in the 20th century.

History

The first time CPR was heard of, was in a biblical story about the prophet Elijah. The story mentioned that Elijah was involved in bringing a child back to life. It tells of a mother who brought her child to Elijah. The child was so sick that he was not breathing and the woman asked Elijah for help. He carried the boy to a bed and prayed to the Lord. While doing this he stretched out himself on top of the boy three times. Then, the Lord heard Elijah’s voice and the boy came back to life. There is also another biblical story about the disciple of Elijah, Elisha. In this story, Elisha places himself over a boy with his mouth on his mouth, his eyes on his eyes, and his hands on his hands. The boys body became warm and he stepped down, walked up and down the room, then he got on top of the boy and bent over him. At this point, the boy sneezed seven times and he awakened and opened his eyes. It is said that the weight of Elisha compressed the child’s chest and that Elisha’s beard tickled the boys nose which caused him to sneeze. This was also known as the origin to the phrase, God bless you. In the early ages, different techniques were adopted that were the first attempts at resuscitation. One method was the Flagellation Method that was used in the Early Ages. This method involved a person actually whipping the victim to try and stimulate some type of response. Another method that was used was the Inversion Method. This was used in 1770. This method involved hanging the victim from his feet, with chest pressure to aid in exhaling and pressure release to help with inhaling. This technique was used in Egypt for 3,500 years and became popular in Europe. Another method was the Barrel Method that was used in 1773. This method involved the use of a wooden barrel to force air in and out of the victim’s chest. A person would put the victim on the barrel and roll him or her back and forth which would result in forcing air in and out of the chest cavity. One more method that was unique is the Trotting Horse Method. This was used in 1812 by lifeguards that had horses. They would rescue the victim, then place them on top of the horse and run the horse up and down the beach. This was suppose to result in an alternate compression and relaxation of the chest cavity from the bouncing of the body on the horse. This technique was banned in the United States in1815 because of several complaints from the Citizens for Clean Beaches. Some highlights in history on CPR are: In 1740, the Paris Academy of Sciences officially recommended mouth-to-mouth resuscitation for drowning victims. In 1891, Dr. Friedrich Maass performed the first equivocally documented chest compression in humans. In 1903, Dr. George Crile reported the first successful use of external chest compressions in human resuscitation and in 1904, he performed the first American case of closed-chest cardiac massage.

In the 18th century, the first city to teach and promote resuscitation was Amsterdam. In August of 1767, the Society for Recovery of Drowned Persons formed and it was the first organized effort to respond to sudden death. Their methods to stimulate the body involved several techniques. They recommended to first warm the victim, positioning the victim’s head lower than feet to remove water, applying pressure to the abdomen. Next they recommended respiration into the victim’s mouth, tickling the victim’s throat, stimulating the victim with rectal and oral fumigation using tobacco smoke, then last, bloodletting. The first four techniques are still in use today. Some highlights in history are: in 1956, Peter Safar and James Elam invented mouth to mouth resuscitation. In 1957, the U.S. Military adopted the method to revive unresponsive victims. In 1960, CPR was developed. In 1972, Leonard Cobb held the worlds first mass citizen training in CPR in Seattle, Washington that was called Medic 2. He helped in the training of over 100.000 people the first two years of the programs.

Conclusion

In conclusion, all of the people who used different CPR techniques had one goal, and that was to save a persons life. CPR techniques have changed and have come a long was since the Early Ages and techniques keep on changing and improving until this day. CPR is an important part of first aid skills and it will continue to save thousands of lives each year.

References

Works Cited
for more info and practice
History of CPR. Wikipedia, The Free Encyclopedia. 6 April. 2008. http://en.wikipedia.org/wiki/History_of_CPR

History of CPR. UKDivers. 6 April. 2008. http://www.ukdivers.net/history/cpr.htm

History of CPR. American Hear Association. 6 April. 2008.

sop Samples

2009-07-03T13:22:00.000-07:00

This statement of purpose sample not standard operating procedure

This SOP statement of purpose sample is written for the admission of university is given to you to write your own copy
This describes why you want to join the university

Name: Your name

Field: Masters in Computer Science

Semester Applied: Spring 2002

________________________________________________________________________

Most of the achievers irrespective of their field of excellence, whether, musicians, politicians or managers would certify that they had dream, and that they had pursued their goals relentlessly. My aspiration to climb the ladder is based on the similar lines.

The role of computers in our day to day life has acquired prominence beyond imagination. Computers have virtually invaded every part of our lives and their far-reaching implications need no explanations. The role of computers in diverse fields is spellbinding. The various software developed has grown to meet the demands of almost any kind of user. Yet, there are more surprises to follow.

Thus, in this complex world of continuous development, computers in particular, it is imperative to be competitive and one’s academic qualifications should commensurate with ones imperious skills and concepts.

Hence my goal of becoming a Masters in Computer Science in indicated.

The zeal and determination to purse and explore the field of computer is reflected from my previous qualifications. I have completed my Bachelor Degree in Engineering from Gulbarga University, Karnataka, India, in the field of Computer Science securing a first class with distinction. During the course of my study, we dealt with a wide verity of subjects, mostly on the fundamental aspects of each subject.

An advanced degree in Computer Science will be the line of demarcation of my pursuit for academic excellence. I am sure that it will give me the right impetus and required acumen for reaching the pinnacle of my career in future. And, of course, nothing can be more honorable then to be a part of your esteemed institute, the benefits of which I shall ream and relish in my career ahead.

I am highly motivated, excited with enthusiasm and desirous to join for further studies in your institution, which is the path leading to my goal. I know, given a chance I will be able to learn, apply and contribute to the mankind using computer science and thereby achieve my long-standing ambitions.

I look forward to a happy and everlasting association with your famed institution.

your name

I hope that this helped you in writing it
I will try to provide more samples Thanks for reading this SOP statement of purpose

True False biology questions

2009-06-27T12:18:00.000-07:00

True False biology questions

Write A if true and Write B if False

___The Origin of Species was published soon after the H.M.S. Beagle returned to England.

___The major difference between Darwin and Wallace's views on evolution focused on the relative importance of natural selection as an evolutionary force.

___The "New Synthesis" or "Modern Synthesis" refers to the merging of mutation theory with Darwinian natural selection in the early 1900's.

___Mendelian principles of genetics explain all the sources of heritable variations in natural populations of organisms.

___Linked gene refers to movement of alleles.

___Linkage of gene and crossing-over tend to accelerate evolutionary change.

___Individual organisms do not evolve

___In order to avoid offending religious leaders of the time, Darwin was very careful not to mention "special creation" or the "Creator" in his book.

___Gregor Mendel ("The Father of Genetics) had the opportunity to read Darwin's writings, but Darwin was unaware of Mendel's genetic discoveries.

___For evolution to occur, changes must be heritable

___Evolution is second only to genetics as "the central organizing principle in all of biology"

___Dispersal of an individual from one population to another does not necessarily result in gene flow.

___Charles Darwin was unaware of Alfred Russel ___Wallace's studies until long after the voyage of the Beagle.

___Charles Darwin believed that natural selection was a very powerful force and that every part of an organism was "designed" by natural selection to serve a particular purpose.

___"Point" or "gene" mutations are mutations that occur at weak points in chromosomes.

End of biology questions

biology multiple choice test

2009-06-24T06:04:00.000-07:00

In this Biology multiple choice test

CHOOSE THE BEST ANSWER. Write only the corresponding letter of your choice on the answer sheet.

1_ What causes water molecules to adhere and cohere to each other to form a column when poured?
1. polarity
2. hydrogen bonding
3. covalent bonding
4. neutral charges
5. all of the above

2. The epidermis of a green stem of an herbaceous plant contains numerous small green spheres called
1. chloroplasts
2. mitochondria
3. guard cells
4. nuclei
5. vacuole

3. The process by which chloroplasts moving about the perimeter of a live cell is termed
1. vibrating cells
2. cyclosis or cytoplasmic streaming
3. high power magnification
4. epidermal streaming
5. photosynthesis

4. Which of the following processes you used for separation of organelles in carrot root cells?
1. filtration
2. centrifugation
3. grinding and homogenization
4. staining
5. A, B, and C are correct

5. Which of the following can be observed in the sediment after filtration of ground carrot root with sucrose-buffer solution
1. whole or fragmented cells
2. starch grains
3. vessel elements
4. oxalate crystals
5. all of the above

6. What was the purpose of adding iodine potassium iodide (I2KI) when you examined carrot cells?
1. to stain the cell walls
2. to stain the starch granules
3. to separate the chloroplasts
4. to separate the chromoplasts
5. to precipitate the organelles

7. After centrifugation of homogenized carrot roots, the orange-tinted liquid on top is termed
1. sediment
2. residue
3. supernatant
4. filtrate
5. cytosol

8. Different cellular materials/organelles are present after filtration and centrifugation because cellular parts vary in
1. density
2. sizes 3. solubility
4. all of the above
5. both A and B only

9. Carrot roots homogenized and centrifuged with sucrose-buffer solution produce colored supernatant although chromoplasts remain intact because carrot cells
1. have cytoplasmic orange pigments
2. have water- soluble carotenoid pigments
3. have cytosol that is colored orange
4. have orange cell walls
5. have chromoplasts that do not settle

10. What two structures are shown by pith cells of the stem when observed under a compound microscope?
1. cell wall and nucleus
2. cell wall and mitochondria
3. cell wall and chloroplasts
4. cell wall and chromoplasts
5. cell wall and cytoplasm

11. When a glass rod is rubbed with a piece of cloth and placed near a stream of water,
1. water stream repels it
2. water stream stops
3. water stream is attracted to it
4. water stream flow hastens of slows down
5. glass rod is charged

12. Which is the correct structural formula of water illustrating its polarity?
1. H2O
2. O-

H H

3. H - O- H
4. O – H – O
5. H+

O O

13. A drop of water compared to a drop of ethyl alcohol, water has
1. higher elevation
2. higher rate of evaporation
3. lesser diameter
4. smooth margin
5. all of the above except B

14. What causes water molecules to adhere and cohere to each other to form a column when poured?
1. polarity
2. hydrogen bonding
3. covalent bonding
4. neutral charges
5. all of the above

More of biology multiple choice is comming see all the posts for more

Download

setw example

2009-05-25T13:28:00.000-07:00

In this vc++ set example

start new cpp file and name it fano

copy the code below and see the output

#include
#include
using namespace std;

int main () {

int Numa = 1234567890 ;
int NumB = 5432 ;int NumC =2324234 ;
cout << "Numa = " <<><
cout << "Numa = " <<><
<< numb ="\n"><<< endl ;
cout<< " NumC\n " <<<
return 0;
}

set different value in the function and see the outcome

This is compiled using VC++ 6 compiler

try different ones

c++ dll example

2009-05-21T12:58:00.000-07:00

In this c++ dll example we used a dlll to perform simple mathematical operation which is to convert from Celsius to Fahrenheit and vice versa

The first step is creating the dll
1 - Start new Win32 Dynamic link library and name it famo chose simple dll project the middle one
2- Open the code and add the following
__declspec(dllexport)void Fehrenhite_Celcius_converter(int n_operation, double input , double &result) ;
// we used reference here since we want to change the value of output inside the function
below #include "stdafx.h"

and at the buttom
add the following

void Fehrenhite_Celcius_converter(int n_operation, double input , double &result)
{ switch (n_operation) {
case 0 :
result= 1.8 * input +32 ;
break ;
case 1 :
result = ( ( double) 5/ (double) 9) * (input -32) ;
break ;
}
// casting needed here bec 5/9 is considered 0
break ; }
}

compile the dlll

Now start new MFC name it mydll_program dialog application and chose not to select the about box and name iy Celcius_Fehrenhite converter

3- Add the 2 static controls and 2 Edit controls a button , group box and 2 radio buttons as shown above

In the group box name it converting and in the static boxes in the properties chose caption input and for the second one chose caption output

Add member variables in the class wizards m_input for the first edit box and m_outputs for the second edit box both of type CString
Add a variable of type int integer for the first radio button m_operation
in the properties chose group

4- after compiling the dlls copy the lib and dll to dialog directory

5- In the project menu Add to projects then chose files as shown here

In the class wizard add member function for the calculate button and name it On_Conversion or anything

Add the following line at the top of the dialog class
__declspec(dllimport)void Fehrenhite_Celcius_converter(int n_operation, double input , double &result) ;
To include the dlls

put in the On_Conversion add follows

void CMydll_programDlg::On_Conversion()
{
UpdateData();
double inputs , outputs ;
inputs = atof(m_input);
// temperature input
Fehrenhite_Celcius_converter(m_operation , inputs ,outputs) ;
// m_operation represents the type of conversion
m_outputs.Format("%2.3f", outputs);
UpdateData(FALSE);
}

build and compile your program

This lesson teaches you how to make simple static dll application for more see other links in the blog

double action mfc c++ button

2009-05-19T02:43:00.000-07:00

in this double action mfc c++ button tutorial we created twofold hover button with 2 tooltips

Depending on the location of the mouse cursor it performs different action whether it is addition or subtraction

DOWLOADS

This tutorial use reflected message to handle the onclicked message since it is a custom control the mouse clicked can not be handled in the dialog class

It is the hardest example I have ever made

To start the application

1- Start new MFC exe application and name it Hashi or else accept default settings
2- In the resource View make add the following bitmaps

and name them as IDB_BITMAP_PLUS_DOWN , IDB_MINUS_HOVER , IDB_MINUS_DOWN , IDB_LUS_HOVER ,IDB_BITMAP_PLUS_DOWN in order ;
make of dimension 200 ,60 pixels do u better if you have

3- Insert new class from insert menu and name it C_Double_task derived from CButton
4-Add the following data members
CBitmap m_bmp_normal_pic ,m_bmp_subtract_hover ,m_bitmap_subtract_down,m_bt_addition_hover ,m_bmp_addition_pushed;
enum Btn_Condition{ NORMAL ,MINUS_HOVER ,MINUS_DOWN,PLUS_OVER,PLUS_DOWN} ;
Btn_Condition m_previous_Mode ,m_present_mode ;
// five bitmaps and enumerators for statuses

5-Add new constructor ( five parameters ) as follows

C_Double_Task::C_Double_Task(int nornel, int negative_over, int neg_dowm, int plus_hovrt, int plus_pressed)
{

m_present_mode =NORMAL ;
/

m_bmp_normal_pic.LoadBitmap(nornel) ;
m_bmp_subtract_hover.LoadBitmap(negative_over) ;
m_bitmap_subtract_down.LoadBitmap(neg_dowm) ;
m_bt_addition_hover.LoadBitmap(plus_hovrt);
m_bmp_addition_pushed.LoadBitmap(plus_pressed) ;
// loading pictures
}

5-Add the following message handlers and virtual functions

first with DrawItem virtual function

void C_Double_Task::DrawItem(LPDRAWITEMSTRUCT lp_Draw_Items_Struct)
{
CDC *pMyDC =CDC::FromHandle(lp_Draw_Items_Struct->hDC);
CDC memDC_item ;
memDC_item.CreateCompatibleDC(pMyDC);
CBitmap* p_bitmap_mfc ;
switch(m_present_mode) {
// or use if statements
case NORMAL :
p_bitmap_mfc = memDC_item.SelectObject( &m_bmp_normal_pic);
break ;
case MINUS_HOVER :
p_bitmap_mfc = memDC_item.SelectObject( &m_bmp_subtract_hover);
break ;
case MINUS_DOWN :
p_bitmap_mfc = memDC_item.SelectObject( &m_bitmap_subtract_down);
break ;
case PLUS_OVER :
p_bitmap_mfc = memDC_item.SelectObject( &m_bt_addition_hover);
break ;
case PLUS_DOWN :
p_bitmap_mfc = memDC_item.SelectObject( &m_bmp_addition_pushed);
break ;
}
pMyDC->BitBlt( 0,0, lp_Draw_Items_Struct->rcItem.right -lp_Draw_Items_Struct->rcItem.left ,lp_Draw_Items_Struct->rcItem.bottom -lp_Draw_Items_Struct->rcItem.top ,&memDC_item ,0,0,SRCCOPY); memDC_item.SelectObject(p_bitmap_mfc) ; memDC_item.DeleteDC(); }

void C_Double_Task::OnLButtonDown(UINT nFlags, CPoint point)
{
SetCapture();
CRect ButtonRect ;
GetClientRect(ButtonRect) ;
m_previous_Mode =m_present_mode ;
m_present_mode =PLUS_DOWN ;
if(point.x > ButtonRect.left +ButtonRect.Width()/2 {
m_previous_Mode =m_present_mode ;
m_present_mode =MINUS_DOWN ;
}
CButton::OnLButtonDown(nFlags, point); }

void C_Double_Task::OnLButtonUp(UINT nFlags, CPoint point)
{
CRect upRect ; GetClientRect(upRect) ;
CRect rect_left = upRect ;
rect_left .right = upRect.right/2 ;
if(PtInRect(upRect,point)) {
if(PtInRect(rect_left,point )) {
m_previous_Mode =m_present_mode ;
m_present_mode =PLUS_OVER ;
ReleaseCapture();
}
else
{ m_previous_Mode =m_present_mode ;
m_present_mode =MINUS_HOVER ;
ReleaseCapture(); }
}
else {
m_previous_Mode =m_present_mode ;
m_present_mode =NORMAL ;
ReleaseCapture();
}
CButton::OnLButtonUp(nFlags, point); }
void C_Double_Task::OnMouseMove(UINT nFlags, CPoint point)
{
CRect rec_total ;
GetClientRect(rec_total) ;
CRect Rec_right = rec_total ;
Rec_right.left = rec_total.right/2 ;
CRect Rct_left = rec_total; Rct_left.right =rec_total.right/2 ;
if( PtInRect( rec_total ,point )){
if( PtInRect( Rec_right ,point )){
m_previous_Mode =m_present_mode ;
m_present_mode =MINUS_HOVER ;
Invalidate(FALSE) ; } else
{ m_previous_Mode =m_present_mode ;
m_present_mode =PLUS_OVER ;
Invalidate(FALSE) ;
} } else if((m_previous_Mode == PLUS_DOWN || m_previous_Mode ==MINUS_DOWN) && PtInRect(Rec_right ,point)){
m_previous_Mode =m_present_mode ;
m_present_mode =MINUS_DOWN ; } else if((m_previous_Mode == PLUS_DOWN || m_previous_Mode ==MINUS_DOWN) && PtInRect(Rct_left ,point)){ m_previous_Mode =m_present_mode ; m_present_mode =PLUS_DOWN ; } else{ m_previous_Mode =m_present_mode ; m_present_mode =NORMAL ; Invalidate(FALSE) ; } TRACKMOUSEEVENT tn_Events ;
tn_Events.cbSize =sizeof(TRACKMOUSEEVENT) ;
tn_Events.dwFlags = TME_LEAVE ;
tn_Events.hwndTrack= m_hWnd ;
_TrackMouseEvent(&tn_Events) ;
CButton::OnMouseMove(nFlags, point); }

void C_Double_Task::OnMouseLeave() {
m_previous_Mode =m_present_mode ;
m_present_mode =NORMAL ;
Invalidate() ; }

and finally add onclicked
afx_msg void OnClicked(); in class header file

ON_CONTROL_REFLECT(BN_CLICKED, OnClicked)

in the begin message map section and add this to your cpp file void C_Double_Task::OnClicked() { DWORD dwpos = GetMessagePos() ;
CPoint pt ;
pt.x = LOWORD(dwpos );
pt.y = HIWORD(dwpos) ;
CRect client_rectangle ;
GetClientRect( client_rectangle) ;
CString strA , strB ;
double db_a ,db_b ;
GetParent()->GetDlgItemText(IDC_NUM1,strA) ; GetParent()->GetDlgItemText(IDC_NUM2,strB) ;
db_a = atof(strA);
db_b= atof(strB);
double result ;
result = db_b + db_a ;
CRect leftRect ; leftRect = client_rectangle ;
leftRect.right = client_rectangle .left +client_rectangle.Width()/2;
if(leftRect.PtInRect(pt)) result = db_b - db_a ;
CString Str_result ;
Str_result.Format("%3.2f",result);
GetParent()->SendMessage(UDM_REGION_CLICKED, result);
}
lastly add this to your class header ;
static const UINT UDM_REGION_CLICKED = ::RegisterWindowMessage(_T("UDM_REGION_CLICKED-{23C118F8-01}"));

now we finished with the custom class Derived from Cbutton

Now we are in the dialog class
6- In the class wizard add the following data member
7- edit the constructor as follows
CHashiDlg::CHashiDlg(CWnd* pParent /*=NULL*/) : CDialog(CHashiDlg::IDD, pParent),m_two_actions_botn(IDB_NORMAL ,IDB_MINUS_HOVER ,IDB_MINUS_DOWN , IDB_LUS_HOVER , IDB_BITMAP_PLUS_DOWN)
{ //{{AFX_DATA_INIT(CHashiDlg) m_numB = _T(""); //}}AFX_DATA_INIT
// Note that LoadIcon does not require a subsequent DestroyIcon in Win32 m_hIcon = AfxGetApp()->LoadIcon(IDR_MAINFRAME); }

8-as pretranslate message handler as follows
C_Double_Task m_two_actions_botn;
and for edit controls add member variable with names like m_numA , m_numB ,m_results of type string

in the dialog class header add afx_msg LRESULT OnRegionClicked( WPARAM wParam, LPARAM lParam );

in the begin message map section in the cpp file ON_REGISTERED_MESSAGE(UDM_REGION_CLICKED, OnRegionClicked)
and at the end of the cpp add
LRESULT CHashiDlg::OnRegionClicked( WPARAM wParam, LPARAM param ) {
CString resl ;
resl.Format("%3.3",wParam);
SetDlgItemText( IDC_RESULT , resl) ;
return (0);
}

finally build the application and trying clicking on the right half and left half
BOOL CHashiDlg::PreTranslateMessage(MSG* pMsg) {
if( m_vc_tooltip.m_hWnd) m_vc_tooltip.RelayEvent(pMsg) ;
return CDialog::PreTranslateMessage(pMsg); }

Note this is a draft not working

c++ hover button with two tooltips

2009-05-09T04:12:00.001-07:00

In this c++ hover button with two tooltips tutorial we create a c++ button control with two tooltips one on the left half and the other on the right side

The tooltips text on the leftis different from one on the right and change in the event of mouse move in the mouse rectangle

To Get the project along the exe click here note that exe is in the debug folder

With the help of codeproject I made this application

To Create the application

1- Start new MFC appWizard( Exe) Dialog based application and give it a name like Tache or any other weird name you want !!?

2- In The Resource editor add the follow images for hover animation

and name them as follows IDB_NORMAL ,IDB_OVER and IDB_DOWE respectively ( the ids of the bitmaps

3-Insert new class in the insert menu or from class view and name it CoolButton
4- Add the following data members to the newly created class ( public)
CBitmap m_bm_Normal ,m_bmp_over ,m_bmp_pushed ;
// three bitmaps and status enumerators
enum Buttom_status { NORMAL ,Hoverimage ,CLICKED} ;
Buttom_status m_previous_state ,m_current_status ;
5- Add new constructor in the new class and edit it as follows

CoolButton::CoolButton(int regular, int hover, int clicjed)
{

m_current_status = NORMAL ;

m_bm_Normal.LoadBitmap(regular) ,
m_bmp_over.LoadBitmap(hover) ,
m_bmp_pushed.LoadBitmap(clicjed) ;
// associate each variable ( Cbitmap with a bitmap )
}

6- Add the following windows messege handlers and virtual functions

Onleftmousedown , onleftmouseup , onmouse_move and Drawitem and edit them as follows

void CoolButton::OnLButtonDown(UINT nFlags, CPoint point)
{

m_previous_state = m_current_status ;
m_current_status = CLICKED ;
// now the button is clicked
SetCapture();

CButton::OnLButtonDown(nFlags, point);
}

and
void CoolButton::OnLButtonUp(UINT nFlags, CPoint point)
{

CRect totals_Rect ;

GetClientRect(totals_Rect) ;

if(On_Recto(totals_Rect ,point))
{

m_previous_state = m_current_status ;
m_current_status = Hoverimage ;
ReleaseCapture();

// if you release the left button while you are inside the button boundaries
}
else {
m_previous_state = m_current_status ;
m_current_status = NORMAL ;
ReleaseCapture();

}

CButton::OnLButtonUp(nFlags, point);
}

void CoolButton::OnMouseMove(UINT nFlags, CPoint point)
{

CRect mybutton_rect ;
GetClientRect(mybutton_rect) ;

if( m_previous_state == CLICKED && (On_Recto(mybutton_rect ,point)))

{

m_previous_state = m_current_status ;
m_current_status = CLICKED ;

}

else if(On_Recto(mybutton_rect ,point))

{

m_previous_state = m_current_status ;
m_current_status = Hoverimage ;
Invalidate(FALSE) ;

}

else
{
m_previous_state = m_current_status ;
m_current_status = NORMAL ;
Invalidate(FALSE) ;
}

TRACKMOUSEEVENT TME_STRUCT ;
TME_STRUCT.cbSize = sizeof(TRACKMOUSEEVENT) ;
TME_STRUCT.dwFlags = TME_LEAVE ;
TME_STRUCT.hwndTrack =m_hWnd ;
_TrackMouseEvent(&TME_STRUCT) ;

CButton::OnMouseMove(nFlags, point);
}

To draw the button
void CoolButton::DrawItem(LPDRAWITEMSTRUCT l_DrawItem_Struct)
{

CDC *pDC_Own = CDC::FromHandle(l_DrawItem_Struct->hDC) ;
CBitmap *pOldBitmap ;
CDC memi_DC
; memi_DC.CreateCompatibleDC (pDC_Own) ;

switch(m_current_status)
{
case NORMAL :
pOldBitmap = memi_DC.SelectObject( &m_bm_Normal ) ;
break ;
case Hoverimage :

pOldBitmap = memi_DC.SelectObject(& m_bmp_over ) ;
break ;

case CLICKED :

pOldBitmap = memi_DC.SelectObject( &m_bmp_pushed ) ;
break ;

}

pDC_Own->BitBlt(0,0, l_DrawItem_Struct->rcItem.right -l_DrawItem_Struct->rcItem.left ,l_DrawItem_Struct->rcItem.bottom- l_DrawItem_Struct->rcItem.top ,&memi_DC ,0,0,SRCCOPY);

memi_DC.SelectObject(pOldBitmap) ;

memi_DC.DeleteDC() ;

}
7- Add the following functioon to Coolbutton class name it On_Recto of type BOOL edit as follows
BOOL CoolButton::On_Recto(CRect &rct, CPoint cpt)
{

return(cpt.x> 0 && cpt.x< (rct.right -rct.left) && cpt.y>0 && cpt.y< (rct.bottom -rct.top) )?TRUE :FALSE ;

to check whether the mouse inside the botton rect
}

8- Add the following virtual function to the new class of type void
void CoolButton::OnMouseLeave()
{

m_previous_state = m_current_status ;
m_current_status = NORMAL ;

Invalidate() ;

}
and add ON_MESSAGE(WM_MOUSELEAVE,OnMouseLeave) inside begin messege map

Now we are finished from CoolButton Class

9- In the dialog resource editor set Ok button properties as owner_draw
10 Open the class wizard and add a member variable chose IDOK in the dialog class and let us name it m_double_tooltip_button or something else and do not forget to put
#include "CoolButton.h" in your dialog class header file

11- Replace the old constructor in the dialgo class with the following one
CTasheDlg::CTasheDlg(): CDialog(IDD_TASHE_DIALOG,NULL), m_double_tooltips_button(IDB_NORMAL ,IDB_OVER ,IDB_DOWE)
{
//loading the bitmaps
}

12- Add the CToolTipCtrl variable in the dialog class and name it like m_mytooltip or else

add the following code to OnItDialog
SetWindowText("helo");
UpdateData(FALSE);
m_double_tooltips_button.MoveWindow( 60,60,240,140) ;
// buton new position
CRect button_rectangle ;
m_double_tooltips_button.GetClientRect(button_rectangle) ;

m_mytooltip.Create( this);

CRect leftRect = button_rectangle ;
CRect Right_button = button_rectangle ;

leftRect.right = Right_button.left = button_rectangle.right/2 ;
m_mytooltip.AddTool(&m_double_tooltips_button, " Right half ", &Right_button ,1) ;
m_mytooltip.AddTool(&m_double_tooltips_button, " you are on the left part of the custom control botton",&leftRect ,2) ;
// or you can replace last addtool with updateTooltip function

13 - In your dialog class add the PreTranslateMessage messege handler as follows

BOOL CTasheDlg::PreTranslateMessage(MSG* pMsg)
{

if(m_mytooltip.m_hWnd) m_mytooltip.RelayEvent(pMsg) ;

return CDialog::PreTranslateMessage(pMsg);
}

14- Build application compile it and move the mouse cursor inside the botton and see the effect of that

I hope that you liked this tutorial ( tache) for more tutorials see the blog for more

My next project would be A button with 2 functions

Thanks for reading the tache program ....

c++ Hover button with two state images left and right

2009-04-22T13:11:00.000-07:00

In this c++ example hover button with two state images is based on the example
http://www.codeproject.com/KB/buttons/Cool_Buttons.aspx in codeproject.com

I deleted unnecessary codes and I added my stuff to it
I could not use SetStatus function so I wrote it every time I use it( 2 line functions)
Every thing that does not make changes I omitted it

DOWNLOAD The downloads contains complete file plus the exe in the folder of debug

The idea and the theme of this project is to have a button with two state hover
like the one shown in the images below,

The first image is when the mouse cursor on the right side and the second one is when the cursor is over the left side

These depend on the location of mouse cursor

To expand this project you can make multifunction buttons or double function buttons

or multiple hover button or another thing to add is to use 2 tooltips for a single button

1-Start visualc++ project MFC Application wizard let name a weird name tasho or tashi if you like and Chose Dialog and Deselct about Box and activeX controls ( no need for them)
and deselect compiler comments generated by the compiler

2-Rename the Dialog resource ID to IDD_TASHO_DIALOG

3-Add the following bitmap resources to resources go to resource View and Insert Insert Bitmap
and draw 5 bitmaps as shown here

anyway Send me better if you can

Make the images of size 40 120

chose id for the bitmaps IDB_NORMAL ,IDB_LEFT ,IDB_RIGHT ,IDB CLICKED ,IDB_DOWN
note that IDB CLICKED is currently unused but it will be used in future programs
4- In the insert menu Insert new class MFC class Chose Cbutton as the base class for it and Name it CCoolButton
5- Add the following data members to the newly created class
and make them public

CBitmap m_bmp_normal ,m_bmp_left ,m_bmp_right ,m_bmp_clicked ,m_bmp_down;
enum ButtonStatus { NORMAL ,LEFT ,RIGHT ,CLICKED ,DOWN};
ButtonStatus m_old_status ,m_current_status ;

6-In the newly created class
Add the following constructor . five elements
as follows '
CCoolButton::CCoolButton(int nidNormal, int nidleft, int nidright, int nidclicked, int niddown)
{
m_current_status = NORMAL ;

m_bmp_normal.LoadBitmap(nidNormal);
m_bmp_left.LoadBitmap(nidleft);
m_bmp_right.LoadBitmap(nidright);
m_bmp_clicked.LoadBitmap(nidclicked);
m_bmp_down.LoadBitmap(niddown);

}

7- Add the 2 windows messege handler and edit them as follows
first OnleftbuttonDown

void CCoolButton::OnLButtonDown(UINT nFlags, CPoint point)
{
m_old_status = m_current_status ;
m_current_status = DOWN ;
SetCapture();

CButton::OnLButtonDown(nFlags, point);
}

and OnleftbuttonUp

void CCoolButton::OnLButtonUp(UINT nFlags, CPoint point)
{

CRect totalrect ;
GetClientRect(&totalrect);
CRect left = totalrect;
left.DeflateRect(60,0);
if( PtInRect(totalrect,point))
{
if(PtInRect(left ,point))
{
m_old_status = m_current_status ;

m_current_status = RIGHT ;
ReleaseCapture();
}
else {
m_old_status = m_current_status ;

m_current_status = LEFT ;
ReleaseCapture();

}}
else
m_old_status = m_current_status ;

m_current_status = NORMAL ;
ReleaseCapture();

CButton::OnLButtonUp(nFlags, point);
}

and OnMouseMove as follows

void CCoolButton::OnMouseMove(UINT nFlags, CPoint pnt)
{

CRect buttonrect , leftrect ;

GetWindowRect(buttonrect);
leftrect = buttonrect;
leftrect.DeflateRect(60 ,0);

if (m_current_status ==DOWN )
{

if( sRect(buttonrect ,pnt ))
{
Invalidate(FALSE) ; }

else{

m_old_status = m_current_status ;
m_current_status = CLICKED ;
Invalidate(FALSE) ;
}}

else if (m_old_status == DOWN && (PtInRect(buttonrect,pnt)))
{
m_old_status = m_current_status ;
m_current_status = DOWN ;
}

else {
if(sRect ( buttonrect, pnt ))

{if(sRect(leftrect , pnt))
{
m_old_status =m_current_status ;
m_current_status = LEFT ;
Invalidate(FALSE);
}
else

{

m_old_status =m_current_status ;
m_current_status = RIGHT ;
Invalidate(FALSE);

}
}

else {
m_old_status =m_current_status ;
m_current_status = NORMAL ;
Invalidate(FALSE);

}
}

TRACKMOUSEEVENT tm_Event ;
tm_Event.dwFlags = TME_LEAVE ;
tm_Event.hwndTrack = m_hWnd ;
tm_Event.cbSize =sizeof(TRACKMOUSEEVENT) ;

_TrackMouseEvent(& tm_Event );

CButton::OnMouseMove(nFlags, pnt);

}
As you can see I could not use SetStatus in the original example

Also add the virtual function DrawItem

void CCoolButton::DrawItem(LPDRAWITEMSTRUCT lp_di_struct)
{

CDC *DrawDc = CDC::FromHandle(lp_di_struct->hDC);
CBitmap * pOldBitmap ;
CDC membtmDC ;
membtmDC.CreateCompatibleDC(DrawDc) ;

switch( m_current_status)
{
case NORMAL :
pOldBitmap = membtmDC.SelectObject( & m_bmp_normal) ;
break ;

case LEFT:
pOldBitmap = membtmDC.SelectObject(& m_bmp_left);
break ;
case RIGHT :

pOldBitmap = membtmDC.SelectObject(& m_bmp_right);
break ;

case DOWN :
pOldBitmap = membtmDC.SelectObject(& m_bmp_down);
break ;

case CLICKED :
pOldBitmap = membtmDC.SelectObject(& m_bmp_clicked);
break ;

}

DrawDc->BitBlt(0,0, lp_di_struct->rcItem.right -lp_di_struct->rcItem.left , lp_di_struct->rcItem.bottom - lp_di_struct->rcItem.top , &membtmDC,0,0,SRCCOPY);

membtmDC.SelectObject(pOldBitmap);

membtmDC.DeleteDC();

}
As you can see it is almost the same as in the original example but there is 5 switch statements instead of 3 because we have here 5 bimaps notice I altered some names

Then add new function and name it sRect of type BOOL
BOOL CCoolButton::sRect(CRect &rct, CPoint &pnt)
{

return(pnt.x> 0 && pnt.x< (rct.right -rct.left) && pnt.y>0 && pnt.y< (rct.bottom -rct.top) )?TRUE :FALSE ; }

And finally add new function OnMouseLeave and edit as follows

void CCoolButton::OnMouseLeave()
{
m_old_status = m_current_status ;

m_current_status = NORMAL ;
Invalidate();

}

Since this function is part of CWnd class add the following to BeginMessageMap
ON_MESSAGE(WM_MOUSELEAVE,OnMouseLeave)
under
ON_WM_LBUTTONDOWN()
ON_WM_LBUTTONUP()
ON_WM_MOUSEMOVE()
in ur class cpp file

Now we finished CCoolButton we move to the dialog class CTashoDlg
class

8-In the dialog resource make In the class wizard Add member variable chose the Ok button chose variable type CCoolButton and let us chose a name for it m_mono or anything you like and do not forget to add #include "CoolButton.h" in your dialog class header file

9-In the InitDialog Add the following code snippet

SetWindowText("hiee");
UpdateData(FALSE);
m_mono.MoveWindow( 30,45,150,90);

10- Delete the original constructor and replace it with the following constructor with no arguments
CTashoDlg::CTashoDlg() :CDialog(IDD_TASHO_DIALOG,NULL), m_mono(IDB_NORMAL ,IDB_LEFT ,IDB_RIGHT ,IDB_CLICKED ,IDB_DOWN)
{
}

11- Build the application and see the result

My next project would be a button with 2 ToolTips one on the left and one on the right I hope that you enjoyed it and benefited from it /.......

for more tutorials see other posts in the blog

It is hard and I made it loll

If you have any comment and any ideas you want to share leave it below :!~

visual C++ Crect InflateRect Example

2009-03-21T03:11:00.000-07:00

In this visual c++ CRect InflateRect example We build simple application that enlarge a rectangle and changing color
To initiate the move you press inside the small rectangle to get irregular enlargement or out side the rectangle to get regular inflation and multiple colors

To the Executable file click here InflatidREct_exe and the complete project theproject

Start Building the project

1- Start new MFC exe single document and name it like inflated_rect
2 - Add the following data members to your derived View class
public:
CRect m_rect_red ,m_rect_blue ;
CBrush m_bar_red ,m_br_green ,m_brush_blue ,m_br_orange ;

and initiliza them in Cview constructor as follows

CInflated_rectangleView::CInflated_rectangleView(): m_rect_red(40,40,120,400),m_rect_blue(70,210,90,230),m_bar_red(RGB(99,0,0)),m_br_green(RGB(0,99,0)),m_brush_blue(RGB(0,0,99)),m_br_orange(RGB(0xCC,0x99,33))
{

}

3- Add the following messege handlers to your derived View class WM_RBUTTONDOWN ,WM_LBUTTONDOWN ,WM_TIMER

and edit the functions they generate as follows

void CInflated_rectangleView::OnLButtonDown(UINT nFlags, CPoint point)
{

if(m_rect_blue.PtInRect(point)){
m_rect_blue.InflateRect(2,5,1,12) ; // irregular inflation
InvalidateRect(NULL,FALSE);
}
else
SetTimer(1, 300,NULL) ; CView::OnLButtonDown(nFlags, point);
}

void CInflated_rectangleView::OnRButtonDown(UINT nFlags, CPoint point)
{
KillTimer(1);
CView::OnRButtonDown(nFlags, point);
}

and finaly

void CInflated_rectangleView::OnTimer(UINT nIDEvent)
{
if( m_rect_blue.left>50)
m_rect_blue.InflateRect(4,1) ;
else
m_rect_blue.InflateRect(0,5);

if(m_rect_blue.top<>
m_rect_blue.SetRect(70,210,90,230); // return to first siza

InvalidateRect(NULL,TRUE);
}

4-

Finally edit t the ondraw function to as follows \

void CInflated_rectangleView::OnDraw(CDC* pDC)
{

pDC->TextOut(40,15," Press the left key to see the effect amd try to press inside the blue");

pDC->SelectObject( m_bar_red) ;
pDC->Rectangle( m_rect_red);
pDC->SelectObject(m_brush_blue) ;
if(m_rect_blue.left<=45)
pDC->SelectObject ( m_br_green);
if(m_rect_blue.top <=120)
pDC->SelectObject( m_br_orange);

pDC->Rectangle( m_rect_blue) ;

}

and then build the applications and enjoy it

Visual c++ IntersectRect example tutorial

2009-03-18T13:37:00.000-07:00

In this visual c++ IntersectRect Example we will learn how to use the mfc intersection function

The intersected rectangle hatched in Red

to get the exefile click here and download the file and get the complete project and thinks of additions you can make

To start

1- Start a new project and let us name it IntersectRectangle

2- chose single document and deselect printing and print view no need for them

3- Add the following members to the view class

public:

BOOL m_b_IsIntersected; // weather there is intersection or not

CRect m_Gray_Rectangle , m_Moving_Rectangle ,m_RectIntersection ;

// constant REct and moving and the rectangle of intersection

make your data members public to avoid using getters and setters

4-Initilize the values of the two rectangles in the constructor as follows

CIntersectrectangleView::CIntersectrectangleView():m_Moving_Rectangle(100,60,160,120),m_Gray_Rectangle(90,80,120,110),m_b_IsIntersected(FALSE)

Add the following messege handler to the derived view class

WM_TIMER ,WM_LBUTONDOWN, WM_RBUTTONDOWN

and edit them as follows

start with Wm timer

void CIntersectrectangleView::OnTimer(UINT nIDEvent) {

if(m_Moving_Rectangle.right> 300) m_Moving_Rectangle.OffsetRect(-280,0) ;

m_Moving_Rectangle.OffsetRect(20,0);InvalidateRect(NULL,TRUE); CView::OnTimer(nIDEvent);}

and

void CIntersectrectangleView::OnRButtonDown(UINT nFlags, CPoint point) { KillTimer(1);

CView::OnRButtonDown(nFlags, point);}

and

void CIntersectrectangleView::OnLButtonDown(UINT nFlags, CPoint point) {

SetTimer(1, 450,NULL) ;

CView::OnLButtonDown(nFlags, point);

}

edit the ondraw member functions as follows

void CIntersectrectangleView::OnDraw(CDC* pDC){

pDC->TextOut(60,13,"Press the right button to start moving the black rectangle and the right to stop it and notice the intersection");

pDC->SelectStockObject(BLACK_BRUSH); 

pDC->Rectangle(m_Moving_Rectangle); 

pDC->SelectStockObject(LTGRAY_BRUSH); 

pDC->Rectangle(m_Gray_Rectangle);

LPRECT lp_move = m_Moving_Rectangle.operator LPRECT() ; 

// cast the moving rect to LPRECT 

LPRECT lp_gray = m_Gray_Rectangle.operator LPRECT() ;

// and so as to the grayrect 

m_b_IsIntersected = m_RectIntersection.IntersectRect( lp_move ,lp_gray);
// find whethere there is intersection and draw the intersection rect with hatchbrush
if( m_b_IsIntersected){ 

CBrush brushHatch(HS_DIAGCROSS, RGB(255, 90, 0)); 

pDC->SelectObject(brushHatch) ; 

pDC->Rectangle(m_RectIntersection);
//I used getstockobject because it is easier to use  

}

8- finally build the application and see how it works

Do not forget to use casting when using intersictRect it only accept arguments of type LPRECT

for more example see the rest of the blog

I hope that you enjoyed this tutorial

}

CFormView Example

2009-02-03T10:48:00.000-08:00

In this Visual c++ CformView Example we used a class derived instead of CDialog
This applications contains 2 buttons one for devision and one for multiplication and it has 2 Edit controls to hold the 2 number involved in the arithmetic operation

After pressing the devision button here what you get

To create the application
1-Start a new MFC application wizard chose a name for the project let us CFORMVIEWEX or any thing u like
2-In the last step chose CFormView as the parent of your derived view class

3-Start building the dialog as shown above chose IDC_EDIT1 ,IDC_EDIT2 For the numbers on the lefts ,IDC_RESULT FOr the result number IDC_MULTIPLY FOR multiplication button finally chose IDC_DEVIDE for the devision

4-In class wizard the following member variables m_number1 for the first edit control NUM1 ;
m_number2 for the second edit control second number and finally for the result m_results do not forget to chose CString type for all of the variables
5- Add messege handler for the 2 buttons and name them Mulitiplication and Devide Respectively and edit the code of those functions

Here what you get the code is marked in Red

void CCFormViewExView::Mulitiplication()
{

double first_number , second_number , result;

UpdateData();// Read data

first_number = atof( m_number1) ;
second_number = atof( m_number2);

result = first_number * second_number ;

m_results.Format("%.3f", result) ;
UpdateData(FALSE); // WriteData

}

and the second function

as follows

void CCFormViewExView::Devide()
{

double NumA ,NumB , results ;

UpdateData();
NumA = atoi( m_number1);
NumB = atoi(m_number2);
results = NumA/NumB ;
m_results.Format("%.5f", results) ;
UpdateData(FALSE);
}

6- Finaly build the application stard putting numbers on the edit control and see the results
note the effect of using different parameters in Format function where Devision has 5 decimal numbers than Multiplication which has only 3

To download the whole project click here fullproject and if you want the exe file click here Exefile

This post contains many typos like devide and others i made them for search engines intentionaly Loll

Microsoft Visual c++ Timer Example

2009-01-28T12:44:00.000-08:00

In This Tutorial microsoft visual c++ Timer example we use left mouse buttons to trigger the motion of a rectangle and kill the motion of a rectangle m_rect_time when left button up

=

note the photos here are protected by copyright rules

DOWNLOADS

The downloadds contain the whole project plus the Exe in the debug folder
To start rect motion presss left mouse button and stop it press right mouse button click

and start again using the left mouse the time was chosen is 300 milli seconds

To start the tutorial

1- Start new single document application and let us name it TimeEX and deselect printing and print proview

2- In your derived view class add the following public data member m_rect_time ;

3- In your view class initilise it as follows :m_rect_time( 20,20,50,50) .

4- Add windows messege handlers for WM_TIMER , WM_RBUTTONDOWN, WM_LBUTTOMDOWN .

5- Edit the OnTimer function as shown in red
// timer action

void CTimerEXView::OnTimer(UINT nIDEvent) {

if(m_rect_time.right> 250 ) m_rect_time.OffsetRect( -250,0) ;

// rectangle motion

m_rect_time.OffsetRect( 50,0) ;

InvalidateRect(NULL,TRUE);

CView::OnTimer(nIDEvent);

}

6-Edit on left button handler as shown here in red

SetTimer(1, 300,NULL) ; // 3 00 millisecond

// trigger the timer by setting the time 3 00 millisecond 1 mearns first timer NULL timers window

7- Edit the on right buttom down as folows

KillTimer(1);

// we have only one timer

8-Edit the ondRaw as follows

pDC->TextOut(70 ,70,"Press the left mouse button to start moving the rectangle using timer and right button to stop it") ;

pDC->SelectStockObject(BLACK_BRUSH );

pDC->Rectangle(m_rect_time) ;

9- Build the application and start stoping the rectangle as what I did here is the figure below

of course you can use bigger font for the text and can change tiime interval by change the value in settimer second paprameter

Press CTrl F5 abd build it

to download the whole project click here
I hope that I guided how to create simple timer tutorial

Hints To practice
if you want multiple timers then is what you have to do is to trigger multiple timer by the function settimer by changing the first and second parameter
for example if you have second timer SetTimer( 2,400, NULL)
second one is set to 400 and add handler to killtimer by killTimer(2) ;
and so on also try different values in the second parameters in settimer ( ) and see the effects

Use callBack functions instead of the WM_TIMER you used and try to control the clock by increasing and decreasing time intervals using menus
for lessons ans extra tutorials see the rest of the blog

for notes and and comments leave me a post below

Creating Dual Timers

1 Create buttons in the toolbar as shown in the figure below

Make like them
Then assign the following ids for the buttons

IDC_GO_LEFT , IDC_BAN_HORIZONTAL_MOVES ,IDC_ACTIVATE_VERTICAL_MOVES AND IDC_FREESE_UPDOWN

2- In the view class add the following member variables
BOOL m_b_halt_leftright,m_Bol_suspended_UpDown_motion
;
CRect m_animated_sqaure ;

Make them public
The Boolean variables are used to enable moving and the square is the animated object

3-Go to the constructor of the class and edit it as follows

CDualtimersView::CDualtimersView():m_animated_sqaure(20,20,50,50)
{

m_b_halt_leftright = TRUE ;

m_Bol_suspended_UpDown_motion = TRUE ;

}

// so at first the square is stopped

Now go on Draw virtual function and edit as follows

void CDualtimersView::OnDraw(CDC* pDC)
{
pDC->TextOut(40,10," Press Toolbars bottons");

pDC->SelectStockObject(BLACK_BRUSH );
pDC->Rectangle(m_animated_sqaure) ;

}

// The ondraw vertual function with dimentions 30 pixels by 30 pixels draw black square

Add windows message handler for the WM_timer and edit as

void CDualtimersView::OnTimer(UINT IDEvent)
{

switch(IDEvent)
{

case 1:

// first timer

if(m_animated_sqaure.right> 250 )
m_animated_sqaure.OffsetRect( -250,0) ;
m_animated_sqaure.OffsetRect( 20,0) ;
InvalidateRect(NULL,TRUE);

break ;
case 2:
// second timer and the action to shift
if(m_animated_sqaure.top <10)
m_animated_sqaure.OffsetRect(0,200);
m_animated_sqaure.OffsetRect( 0,-10) ;
InvalidateRect(NULL,TRUE);
break;
default:

CView::OnTimer(IDEvent);
}

}

Now add windows mesege handler for the Tolbar buttons

We start with the first button

void CDualtimersView::OnGoLeft()
{
// see if the button not pressed
if(m_b_halt_leftright) {

SetTimer( 1,500,NULL) ;
// half a second
m_b_halt_leftright =FALSE ;
}
}

and add Update windows messge handler for the same button
void CDualtimersView::OnUpdateGoLeft(CCmdUI* pCmdUI)
{
pCmdUI->Enable(m_b_halt_leftright);

}

Do the same for the whole buttons as you can see here

void CDualtimersView::OnBanHorizontalMoves()
{
if(!m_b_halt_leftright) {
KillTimer(1);
m_b_halt_leftright =TRUE ;
}

}

void CDualtimersView::OnUpdateBanHorizontalMoves(CCmdUI* pCmdUI)
{
pCmdUI->Enable(!m_b_halt_leftright);

}

void CDualtimersView::OnActivateVerticalStep()
{
if(m_Bol_suspended_UpDown_motion) {
SetTimer( 2,200,NULL) ;
m_Bol_suspended_UpDown_motion =FALSE ;
}

}

void CDualtimersView::OnUpdateActivateVerticalStep(CCmdUI* pCmdUI)
{

pCmdUI->Enable(m_Bol_suspended_UpDown_motion);

}

void CDualtimersView::OnFreezeUpdown()
{

if(!m_Bol_suspended_UpDown_motion) {
KillTimer(2);
m_Bol_suspended_UpDown_motion =TRUE ;
}

}

void CDualtimersView::OnUpdateFreezeUpdown(CCmdUI* pCmdUI)
{
pCmdUI->Enable(!m_Bol_suspended_UpDown_motion);

}

As an activity Add discription to the code i did
Note use better method for naming
and also see the toolbar example for more on toolbars and use menu items to start and stop the motion and add keyboard accelerators

Click here to download both examples Download link

Please leave your feedback below

Microsoft visual c++ toolbar example

2009-01-26T11:12:00.000-08:00

Microsoft visual c++ toolbar example use toolbar buttons to move movable rectangle m_rect_movable in four directions up, down , left and right and when intersected with a fixed rectangle m_rect_fixed it draws new rectangle with red - hatched pattern and also it the movable rectangle become blue hatched when it approaches the border rectangle m_rectBorder .

This program is similar to a previous program that used menu items to control the position of a movable rectangle
There is 2 things to check the first one is to check the intersection between the movable rectangle and the fixed rectangle so that draw a hatched pattern in the intersection rectangle I could not use IntersectRect function because it takes LPCRECT as parameters and the other thing to check was if the movable rectangle is close to the border rectangle ,you can GetClientRect function to get the window's client area instead of the border rectangle

To build the application

1- Chose new and build single doc application and name it like TOOLBAR_MOVE_EX
2- In the resource view go to toolbar and delete some of the default buttoms and add 4 buttoms as shown in the fig below and assign the following IDs to them ID_SHIFT_LEFT ,ID_SHIFT_RIGHT ,ID_MOVE_UP and ID_MOVE_DOWN sorry for the unpretty buttoms !!

3-Add the following data members to your derived CView class ( 3 rectangles and 5 boolean variables)
CRect m_rectBorder , m_rect_movable, m_rect_fixed ; // border ,movable and fixed rectangles respectively
BOOL m_bLeftBorder ,m_bRightBorder,m_bTopBorder ,m_bButtomBorder ,m_bIntersect;
// border and intersection indicators
4- In the view class initilize your data members as follows
CTOOLBAR_MOVE_EXView::CTOOLBAR_MOVE_EXView():m_rectBorder(10 ,10,340,340),m_rect_fixed(100,100,200,200) ,m_rect_movable(40,30,90,130)
{

m_bLeftBorder =m_bRightBorder =m_bTopBorder =m_bButtomBorder =m_bIntersect==FALSE ;

}

5- In your view add windows messege handlers for command and update command as follows

void CTOOLBAR_MOVE_EXView::OnShiftLeft()
{

if(m_rect_movable.left <=(m_rectBorder.left +40)) m_bLeftBorder = TRUE ; m_rect_movable.OffsetRect(-10,0); m_bRightBorder = FALSE ; if (m_rect_fixed.PtInRect(m_rect_movable.TopLeft()) | m_rect_fixed.PtInRect(m_rect_movable.BottomRight())| m_rect_fixed.PtInRect(CPoint(m_rect_movable.right,m_rect_movable.top))|m_rect_fixed.PtInRect(CPoint( m_rect_movable.left,m_rect_movable.bottom))) { m_bIntersect =TRUE ; } InvalidateRect(NULL,TRUE); } void CTOOLBAR_MOVE_EXView::OnUpdateShiftLeft(CCmdUI* pCmdUI) { pCmdUI->Enable(!m_bLeftBorder);

}

void CTOOLBAR_MOVE_EXView::OnShiftRight()
{

if(m_rect_movable.right >= ( m_rectBorder.right -30 ))
m_bRightBorder = TRUE ;

m_rect_movable.OffsetRect(10,0);
m_bLeftBorder = FALSE ;

if (m_rect_fixed.PtInRect(m_rect_movable.TopLeft()) | m_rect_fixed.PtInRect(m_rect_movable.BottomRight())| m_rect_fixed.PtInRect(CPoint(m_rect_movable.right,m_rect_movable.top))|m_rect_fixed.PtInRect(CPoint( m_rect_movable.left,m_rect_movable.bottom))) {
m_bIntersect =TRUE ;

}

InvalidateRect(NULL,TRUE);

}

void CTOOLBAR_MOVE_EXView::OnUpdateShiftRight(CCmdUI* pCmdUI)
{

pCmdUI->Enable(!m_bRightBorder);

}

void CTOOLBAR_MOVE_EXView::OnMoveDown()
{

if( m_rect_movable.bottom >= ( m_rectBorder.bottom -20))
m_bButtomBorder = TRUE ;

m_rect_movable.OffsetRect(0,10) ;
m_bTopBorder = FALSE ;

if (m_rect_fixed.PtInRect(m_rect_movable.TopLeft()) | m_rect_fixed.PtInRect(m_rect_movable.BottomRight())| m_rect_fixed.PtInRect(CPoint(m_rect_movable.right,m_rect_movable.top))|m_rect_fixed.PtInRect(CPoint( m_rect_movable.left,m_rect_movable.bottom))) {
m_bIntersect =TRUE ;

}

InvalidateRect(NULL,TRUE);

}

void CTOOLBAR_MOVE_EXView::OnUpdateMoveDown(CCmdUI* pCmdUI)
{

pCmdUI->Enable(!m_bButtomBorder);

}

void CTOOLBAR_MOVE_EXView::OnMoveUp()
{

if(m_rect_movable.top <= (m_rectBorder.top + 20)) m_bTopBorder =TRUE ; m_rect_movable.OffsetRect(0,-10) ; m_bButtomBorder = FALSE ; if (m_rect_fixed.PtInRect(m_rect_movable.TopLeft()) | m_rect_fixed.PtInRect(m_rect_movable.BottomRight())| m_rect_fixed.PtInRect(CPoint(m_rect_movable.right,m_rect_movable.top))|m_rect_fixed.PtInRect(CPoint( m_rect_movable.left,m_rect_movable.bottom))) { m_bIntersect =TRUE ; } InvalidateRect(NULL,TRUE); } void CTOOLBAR_MOVE_EXView::OnUpdateMoveUp(CCmdUI* pCmdUI) { pCmdUI->Enable(!m_bTopBorder);

}

6- Edit the OnDraw member function of your generated view

pDC->Rectangle(m_rectBorder) ; // white hallow
pDC->SelectStockObject(LTGRAY_BRUSH); // outer rectangle
pDC->Rectangle(m_rect_fixed); // fixed rectangle

CBrush bluebrush(HS_DIAGCROSS, RGB(0, 0, 250)); // blue brush for border condition when moveable rect on border

pDC->SelectStockObject(BLACK_BRUSH);
if(m_bLeftBorder |m_bRightBorder |m_bTopBorder |m_bButtomBorder)
pDC->SelectObject(bluebrush) ; // if in nearby the border rectangle
pDC->Rectangle(m_rect_movable);
if(m_bIntersect) {
CRect rc_intersection ;// use red hatch pattern
rc_intersection.IntersectRect(m_rect_movable ,m_rect_fixed);
CBrush brushHatch(HS_DIAGCROSS, RGB(255, 90, 0));
pDC->SelectObject(brushHatch) ;
pDC->Rectangle(rc_intersection);

}
7- try to move the movable rectangle around when you start the application you will start with a window like this

and after moving it for some bushes on the button you will get similar to this and try to move near the corners to see what happens there

to dowbload the whole project clich the link here

and the exe file

and do not forget to add accelerator use GetClientRect instead of the border Rect

c++ conditional return example

2009-01-25T12:43:00.000-08:00

In this c++ condition return example we made a function the accept both positive and negative numbers and produce the output of the square rout

if the number is negative the output would the square rout of absolute value otherwise the output will the square rout

at first the the program check the number if it is positive or negative

This is simple c++ conditional return example that uses if statement to check the value of the number entered

as seen in the red code below

#include
#include
using namespace std ;

double sqr( double x) {

if( x<0)
return sqrt(-x) ;

return sqrt(x) ;

}

int main() {

cout<< "Enter the number you to get square rout positive or negative "<<>
double x , result ;
cin>> x;
result = sqr (x);
cout<<"The resuly is square rout of the number you entered " <<>

return 0 ;

}

Visual c++ Menu Example

2009-01-23T11:57:00.000-08:00

download
In this Visual c++ menu example we Top level Move with 4 pop_up items Up ,Down ,Left and Right to move a rectangle rcRect inside the borders of a bigger rectangle rcBorder .

working example
To create the program
1- Start new Single document MFC exe application and deselect printing and print view options
2-In the resource View to IDR_MAINFRAME menu and make add the following items as shown in Fig 1

Downloadshttp://www.4shared.com/file/88832447/4d0b1fea/Menu_Move_Rect.html
3- In your derived view class add the following member functions and make them public

BOOL m_bLeft,m_bRight ,m_bDOWN , m_bUp ; // these are the left right top bottom borders indicators

CRect rcRect ,rcBorder ; // small rectangle and the bigger rectangle rcBorder

4-In your derived view class initilise the date as shown
CMove_RectView::CMove_RectView():rcRect(100,100,50,50),rcBorder( 20 ,20 ,300,300)
{

m_bLeft =m_bRight = m_bDOWN =m_bUp =FALSE ;

}

5-Add messege handlers to your menuitem IDs

void CMove_RectView::OnMoveUp() { rcRect.OffsetRect(0 ,-20); m_bDOWN =FALSE ; if((rcRect.top -70) <= rcBorder.top) m_bUp =TRUE ; InvalidateRect(NULL ,TRUE ); } void CMove_RectView::OnUpdateMoveUp(CCmdUI* pCmdUI) { pCmdUI->Enable(!m_bUp); } void CMove_RectView::OnMoveDown() { rcRect.OffsetRect(0, 20); m_bUp =FALSE ; if((rcRect.bottom) >= (rcBorder.bottom -70) ) m_bDOWN =TRUE ; InvalidateRect(NULL ,TRUE ); } void CMove_RectView::OnUpdateMoveDown(CCmdUI* pCmdUI) { pCmdUI->Enable(! m_bDOWN) ; } void CMove_RectView::OnMoveLeft() { if( (rcRect.left) <=rcBorder.left + 70) m_bLeft = TRUE ; rcRect.OffsetRect(-10 , 0); m_bRight = FALSE ; InvalidateRect(NULL ,TRUE ); } void CMove_RectView::OnUpdateMoveLeft(CCmdUI* pCmdUI) { pCmdUI->Enable(!m_bLeft); } void CMove_RectView::OnMoveRight() { rcRect.OffsetRect(20 ,0); m_bLeft = FALSE ; InvalidateRect(NULL ,TRUE ); if((rcBorder.right)<= rcRect.right) m_bRight=TRUE ; } void CMove_RectView::OnUpdateMoveRight(CCmdUI* pCmdUI) { pCmdUI->Enable(!m_bRight); }

6- Finally edit the OnDraw function in your view class as follows

void CMove_RectView::OnDraw(CDC* pDC)
{

pDC->Rectangle(rcBorder) ;
if( m_bLeft |m_bRight |m_bDOWN |m_bUp)
pDC->SelectStockObject(DKGRAY_BRUSH ) ;
// if the small rect near the big rect
pDC->Rectangle(rcRect);

}
7- Build the application and see how it works

8- Try to move the small Rect with the menu options UP DOWN LEFT AND RIGHT And when move near the borders of the big rectangle you get something like in the fig 2

to download the whole project clich here
and the exe file click here

things to add are keyboad accelerators to do

C++ overloaded function example

2009-01-18T11:11:00.001-08:00

C++ overloaded function example in this tuturial we used the c++ property of overloading to use single function for different purposes square rout , addition , multiplication and devision in one single operation

we include cmath library in order to use math functions

The code below marked in red demonstrate how made this sample program

#include
#include
using namespace std ;
double SQUAREROOT_MULTIPLY_SUM_DEVISION( double a){
return sqrt( a);}
double SQUAREROOT_MULTIPLY_SUM_DEVISION( double a , double b){
return a *b ;}
double SQUAREROOT_MULTIPLY_SUM_DEVISION (double a ,double b, double c){
return a + b + c ;}
double SQUAREROOT_MULTIPLY_SUM_DEVISION( double a ,double b, double c , double d){

return a /( b +c +d) ; }
C++ overloaded function example In this tutorial we used the property of overloading to use single function for 4 different purposes Square rout addition multiplication and division

int main (){

cout << " this program used single function to perform square rout , multiplication , addition and devisition in single operation " <<>

double w = SQUAREROOT_MULTIPLY_SUM_DEVISION(4);
cout<< " square root of 4 ="<<>
double x = SQUAREROOT_MULTIPLY_SUM_DEVISION( 12 ,14);
cout << " the sum of 13 and 14 is "<<>
double y = SQUAREROOT_MULTIPLY_SUM_DEVISION( 2 ,3,4);
cout <<" 2 *3*4 = "<<>
double z = SQUAREROOT_MULTIPLY_SUM_DEVISION ( 100, 2,3,5);
cout << " 100 / ( 2 + 3 +5) "<<>
return 0 ;

}

C++ overloaded function example

2009-01-18T11:11:00.000-08:00

MFC visual c++ CstringArray Example

2009-01-17T12:24:00.000-08:00

In this MFC visual c++ CstringArray example we used CstringArray class in order to display time on status bar bans we used it to display seconds , minutes and hours

1 - Build MFC application single document and chose not to select printing and printing preview
2 - Add resource element in the View menu chose resource simbols and New and name it ID_TIMEARRAY
3-Add handlers in the CmainFrame for ID_VIEW_STATUS_BAR and update handler for it,
4-Replace the previously created indicators array with 3 ID_SEPARATOR to hold second minutes and hours respectively and make m_wndStatusBar data member public instead of private .
5- Replace if (!m_wndStatusBar.Create(this) with this, WS_CHILD | WS_VISIBLE | CBRS_BOTTOM, ID_TIMEARRAY) ;so as not to use the default status bar .
6-

void CMainFrame::OnViewStatusBar()
{
 m_wndStatusBar.ShowWindow((m_wndStatusBar.GetStyle() &
WS_VISIBLE) == 0);
RecalcLayout();
}
void CMainFrame::OnUpdateViewStatusBar(CCmdUI* pCmdUI)
{
 pCmdUI->SetCheck((m_wndStatusBar.GetStyle() & WS_VISIBLE) != 0);
} in  mainframe.cpp

7- Add   the following  function OnDraw to your  View class         pDC->TextOut(30, 30 ,Press the left key to see your computer current time on the status bar ");
and add  #include "MainFrm.h" near the top of your view class

8- ADd the following code to OnLbuttonDown handler in your view class
CTime currenttime = GetCurrentTime();
int sec , min , hours ;
hours = currenttime.GetHour();
min = currenttime.GetMinute();
sec = currenttime.GetSecond();
CMainFrame* pFrame = (CMainFrame*) AfxGetApp()->m_pMainWnd;
CStatusBar* pStatus = &pFrame->m_wndStatusBar;

CStringArray timearray ;
timearray.SetSize(3);
CString strsec ,strmin ,strhours ;
strsec.Format("%d ",sec);
strmin.Format("%d " , min);
strhours.Format("%d " ,hours );
timearray[0] = strsec ;
timearray[1] = strmin ;
timearray[2] = strhours ;

pStatus->SetPaneText(0, timearray[0]);
pStatus->SetPaneText(1 ,timearray[1] );
pStatus->SetPaneText(2 ,timearray[2]);

to download the whole project clock here

to download the Exe file click here

visual c++ Status bar example

2009-01-17T08:42:00.000-08:00

1- Make single document MFC application and deselect printing and printview options
2- In the view menu chose resource and then add New resource and then add ID_TIME_STATUSBAR and accept the defaul value (101) .
3- Add message handler and update message handler in the CMainFrame class for both ID_VIEW_STATUS_BAR .
4- Replace the previously created indicators array with single ID_SEPARATOR and make sure that the CMainFrame data member m_wndStatusBar is public instead of protected so that it can be accessed from the View class .
5- Replace if (!m_wndStatusBar.Create(this) with this, WS_CHILD | WS_VISIBLE | CBRS_BOTTOM, ID_TIME_STATUSBAR) ;

6- Add the following code to ur CMainFrame.cpp

void CMainFrame::OnViewStatusBar()
{
m_wndStatusBar.ShowWindow((m_wndStatusBar.GetStyle() &
WS_VISIBLE) == 0);
RecalcLayout();
}
void CMainFrame::OnUpdateViewStatusBar(CCmdUI* pCmdUI)
{
pCmdUI->SetCheck((m_wndStatusBar.GetStyle() & WS_VISIBLE) != 0);
}

7- Add the following function OnDraw to your View class pDC->TextOut(30, 30 ,Press the left key to see your computer current time on the status bar ");

8- Edit OnDraw messege handler as follows

CTime currenttime = GetCurrentTime();
int sec , min , hours ;
hours = currenttime.GetHour();
min = currenttime.GetMinute();
sec = currenttime.GetSecond();
CMainFrame* pFrame = (CMainFrame*) AfxGetApp()->m_pMainWnd;
CStatusBar* pStatus = &pFrame->m_wndStatusBar;
CString strtime ;
strtime.Format( "%d %d %d ", hours,min , sec );
pStatus->SetPaneText(0, strtime );

Finally include #include "MainFrm.h" near the top with other include statements

to download the exe file click here

and the whole project here
this contains sum errors