How c calculates remainder?

Question

I have strange behaviour in one of my functions. Code of program is very huge and hardly readable. So i attaches part of functions. If j lower than 4, it returns not actual remainder, but difference between j and 4.

I tested and found, that 3 % 4 == 1 instead of 3.

while (j < 16)
{
    if ((crd + (j % 4)) + d * (j / 4) > d * d && str2[j] == 1)
        str[0] = 2;
    if (((crd + (j % 4) + d * (j / 4)) / d) - i != (j / 4) && str2[j] == 1)
        str[0] = 2;
    if (str2[j] == 1)
        str[(crd + (j % 4)) + d * (j / 4)] = str2[16];
    if (str[(crd + (j % 4)) + d * (j / 4)] == 2 || str[0] == 2)
        return (str);
    j++;
}

The question is how % operator works.

`3 % 4` is definitely `3`. So, which part of the posted code is misbehaving? And with which data? Post [mcve] please. — Eugene Sh., Jan 31 '19 at 14:40
I'm not sure why you need like 10 lines of codes (that don't even run on their own) for this. `printf("%d %% %d == %d\n", 3, 4, 3 % 4);` does it just as well, and for me the result is `3 % 4 == 3`. Looks like the issue is in your code, not the modulo operation. — Blaze, Jan 31 '19 at 14:40
add some intermediate variables and print them out. I would not trust debugger when there might be some optimized code. — Eugene Sh., Jan 31 '19 at 14:48
@EugeneSh. checked this with gdb and it shows right result. I compiled with gcc -g, so no optimisations were used — Vladimir, Jan 31 '19 at 14:51

score 1 · Accepted Answer · answered Jan 31 '19 at 20:15

the following proposed code:

cleanly compiles
shows that the modulo operator works as expected

and now, the proposed code:

#include <stdio.h>

int main( void )
{
    int a = 3 % 4;
    printf( "%d\n", a );
}

when compiled/linked and run the result is:

above shows that the modulo operator works as expected.

That leaves your code, which is not a [mcve] so the resulting probability is that something is wrong with your code. But we really cannot tell for sure as we don't have your code,

How c calculates remainder?

1 Answers1