Why is integer conversion done this way?

Question

I ran into an interesting scenario with integer conversion:

#include <stdio.h>

int main()
{
    unsigned int x = 20;
    unsigned int y = 40;

    printf("%d\n", x - y);
    printf("%d\n", (x - y) / 4);
}

~ % ./a.out
-20
1073741819

I wasn't expecting the 2nd result. Since x and y are both unsigned ints is the result of x - y unsigned (and in this case displayed as signed by printf)?

You can also try `char * p = x - y;` and look carefully at the error message of the compiler when it complains about an invalid conversion. — David Grayson, Jun 01 '22 at 23:15
The argument corresponding to `%d` should be `int`. To print `unsigned int` use `%u`. — Barmar, Jun 01 '22 at 23:18
The first printf could have shown `-2147483628` if you had one's-complement. It's the conversion from unsigned to signed inside printf that is a problem and implementation defined. — Goswin von Brederlow, Jun 01 '22 at 23:21

score 1 · Answer 1 · answered Jun 01 '22 at 23:17

The things you are printing are indeed unsigned integers and you should print them with %u, but that does not explain the surprising result for the second number. The surprising result comes from an overflow occurring when you calculate x - y, since the result of that subtraction is negative and thus not representable as an unsigned int.

Unsigned overflow/underflow is not undefined behavior, so it's OK to have code like this in production if you know what you are doing.

Why is integer conversion done this way?

1 Answers1