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?
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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?
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:
It outputs 65525Code:#include <stdio.h>
#include <stdint.h>
int main() {
uint16_t a=4; a-=15;
printf("%d\n",a);
return 0;
}
It's undefined behaviour. Best thing to do is check before the subtraction.Quote:
Originally Posted by ridgerunnersjw
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;
}
Ah, right! It's only UB with signed values...Quote:
Originally Posted by john.c