bool

[Chin Pei Ying]

int count=0;
for(i=0; i<size; i++){
for(j=0; j<size; j++){
if(x[i][j] == y[i][j]){
count++;
continue;
}
else
break;}
}
if(count==9)
cout << "Idempotent";
any simple method to check this two matrix?
Please help me
Thank you

[Chin Pei Ying]

#include<iostream>
#include<conio.h>
#define size 3
using namespace std;

void main(){

int a[3][3]={{2,3,4} , {1,2,3}, {4,5.6}};
int b[3][3]= {{2,3,4} , {1,2,3}, {4,5.6}};
bool check (int [][3], int [][3]);
_getch;
}
bool check(int c[][3], int d[][3])
{
bool flag = 0;

for(int i = 0; i < 3; i++)
for(int j=0; j<3; j++) {
if(c[i][j] == d[i][j])
flag = 1;
}

return flag;
}

[Chin Pei Ying]

If i want to use flag,how to use?
What to change in this program

[coder777]

What to change in this program

You're not calling the function check. Instead it is the prototyp. Change it like so:

#include<iostream>
#include<conio.h>
#define size 3
using namespace std;

bool check (int [][3], int [][3]); // Note

void main(){

int a[3][3]={{2,3,4} , {1,2,3}, {4,5.6}};
int b[3][3]= {{2,3,4} , {1,2,3}, {4,5.6}};
if(check (a, b)) // Note
  cout << "identical"; // Note
_getch(); // Note
}
bool check(int c[][3], int d[][3])
{
bool flag = false; // Note

for(int i = 0; i < 3; i++)
for(int j=0; j<3; j++) {
if(c[i][j] == d[i][j])
flag = true; // Note
else // Note
{
  flag = false; // Note
  break; // Note
}
}

return flag;
} 

[Chin Pei Ying]

but if it is not identical,it still cout identical

[coder777]

wait you're right. Do this:

#include<iostream>
#include<conio.h>
#define size 3
using namespace std;

bool check (int [][3], int [][3]); // Note

void main(){

int a[3][3]={{2,3,4} , {1,2,3}, {4,5.6}};
int b[3][3]= {{2,3,4} , {1,2,3}, {4,5.6}};
if(check (a, b)) // Note
  cout << "identical"; // Note
_getch(); // Note
}
bool check(int c[][3], int d[][3])
{
bool flag = false; // Note

for(int i = 0; i < 3; i++)
for(int j=0; j<3; j++) {
if(c[i][j] == d[i][j])
flag = true; // Note
else // Note
  return false; // Note
}

return flag;
} 

[Chin Pei Ying]

Thank you very much,you help me a lot.