Thread: unsigned int rollover?

Standard seems to define unsigned overflow to work as if modulo 1 + the largest value possible for that type. Signed overflow however is undefined behavior.

I believe the value should be 65525 though.

edit: yes, see below:

Code:

#include <stdio.h>
#include <stdint.h>

int main() {
    uint16_t a=4; a-=15;
    printf("%d\n",a);
    return 0;
}

It outputs 65525

Originally Posted by ridgerunnersjw

If I am decrementing an unsigned int (16 bits) by a value say

k -=15;

what happens the next time around if k is sitting at a 4? Seems it would roll over to 65524....Is this correct? Can I keep handling it as an unsigned int?

It's undefined behaviour. Best thing to do is check before the subtraction.

65525, actually, but yes, wraparound is well-defined with unsigned values.

Code:

#include <stdio.h>
#include <stdint.h>
 
int main()
{
    uint16_t n = (uint16_t)(-1); // -1 interpreted as unsigned gives largest value
    printf("%u\n", (unsigned)n); // prints 65535
 
    n = 4;
    n -= 15;
    printf("%u\n", (unsigned)n);      // prints 65525
 
    // The bit pattern can be interpreted as a signed value and gives the
    // correct signed answer due to 2's-complement representation.
    printf("%d\n", (int)(int16_t)n);  // prints -11
 
    return 0;
}

Originally Posted by john.c

65525, actually, but yes, wraparound is well-defined with unsigned values.

Ah, right! It's only UB with signed values...