Question

In this little program:

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

int main() {
    uint8_t a = 0;
    uint8_t b = 255;

    if (a == (b + 1)) {
        printf("Equal\n");
    } else {
        printf("Not equal\n");
    }

    if (a == ((b + 1) & 0xFF)) {
        printf("Equal\n");
    } else {
        printf("Not equal\n");
    }       
}

I get:

Not Equal
Equal

Why doesn’t the comparison work unless I forcibly take the last 8 bits? I’m guessing I’m missing some nuance of unsigned arithmetic here…

I’m using gcc 4.4.5 if that makes a difference.

Answer 1

Because of integer promotions, both operands of the == and + operator are promoted to int.

The expression:

a == (b + 1)

is then equivalent to:

0 == 256

which is false.

The expression: a == (uint8_t) (b + 1) would give you the result you expect (true). Another solution is to use & 0xFF like in your second if statement,