Integers are 4 bytes, pointers are 4 bytes, and doubles are 8 bytes. What would the output be of the following code?

Question

I'm having some issues with this homework question, is the output a constant string of 0's?

#include <stdio.h>

int main()
{
    unsigned char x;
    for (x = 0; x <= 0xFF; x++) {
        printf("0");
    }
    printf("1\n");
}

Undecidable because the size of `char` in the environment is not speficied. — MikeCAT, Feb 23 '21 at 10:02
[Yes on all Posix archs.](https://en.wikipedia.org/wiki/Character_(computing)#char) [No on some others.](https://stackoverflow.com/questions/32091992/is-char-bit-ever-8) — , Feb 23 '21 at 10:11
When exactly did you expect the unsigned char to _not_ have a value between 0 and 0xFF? I mean if this code was written for some obsolete weirdo DSP then the size of char would be the least of your problems... notably, you wouldn't be programming a PC no longer so the rest of main() wouldn't make much sense. — Lundin, Feb 23 '21 at 12:03

score 1 · Answer 1 · answered Feb 23 '21 at 10:13

You can check what will happen by checking the size of the character in bits:

int main(void)
{
    unsigned char x;
    #if CHAR_BIT > 8
    for (x = 0; x <= 0xFF; x++) {
        printf("0");
    }
    printf("1\n");
    #else 
    printf("This loop will never end\n");
    #endif
}

Integers are 4 bytes, pointers are 4 bytes, and doubles are 8 bytes. What would the output be of the following code?

1 Answers1