Thread: help!! problems with if stream

Code:

#include <stdio.h>
int main() {
     int i, j, k, match, tmp[4] = {4, 12, 25, 30}; 
     int num[4][4] = { {34, 55, 125, 55}, {4, 515, 25, 11},{34, 5, 12, 65},{24, 15, 625, 35}};
     
     for(i=0;i<4;i++){
     printf("The matched numbers of row %d are:\n", i+1);
     for(j=0;j<4;j++){
     for(k=0;k<4;k++)
     if(num[i][j]==tmp[k])
        printf("%4d", tmp[k]);
       match = match + 1;
       }
       
if(match = 0)
    
   printf("nil");
   printf("\n");}

     system("pause");
     return 0;
}

the program will find the matched numbers and should show "nil" when any number matched in that array.

the output should be:

Code:

the matched numbers of row 1 are:
nil
the matched numbers of row 2 are:
4  25
...etc

the problem now is that when there isn't any numbers matched, the "nil" doesn't appear....
why?

if(match ==0) !!!

Originally Posted by laserlight

match = 0 probably should be match == 0

Oh, and kindly indent your code properly.

Damn, laserlight you are always sooo fast

wow so fast...

I tried and still it doesn't work.........

I don't even know how your program compiles given the fact that you don't even match brackets. Please post your updated code.

broken brackets?
are there?

Code:

#include <stdio.h>
int main() {
     int i, j, k, match, tmp[4] = {4, 12, 25, 30}; 
     int num[4][4] = { {34, 55, 125, 55}, {4, 515, 25, 11},{34, 5, 12, 65},{24, 15, 625, 35}};
     
     for(i=0;i<4;i++){//first bracket
     printf("The matched numbers of row %d are:\n", i+1);
     for(j=0;j<4;j++){//second brackets
     for(k=0;k<4;k++)
     if(num[i][j]==tmp[k])
        printf("%4d", tmp[k]);
       match = match ++;
       }//second ends here
       
if(match==0)
    
   printf("nil");
   printf("\n");
}//first ends here

     system("pause");
     return 0;
}

thanks, sorry for being a noob