How are the results of intermediate steps in an arithmetic expression stored when all terms in the expression are uint8_t?

Question

The code:

int main(void) {
    uint8_t x = 3;
    uint8_t y = 4;
    uint8_t z = 5;

    uint8_t a = (x - y) / z;
    uint8_t b = x - y;
    printf("a is %d\n", a);
    printf("b is %d\n", b);
    return 0;
}

had the following output:

a is 0
b is 255

I would expect that (3 - 4) / 5 would result in an overflow as (3 - 4) as a uint8_t is 255. When an intermediate step involving unsigned integers results in a negative number, why doesn't it overflow? In what format is the result for (3 - 4) stored before the next step / 5 happens?

Answer 1

One can use a much simpler example to see what's really going on:

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

int main(void) {
    uint8_t x = 3;
    uint8_t y = 4;

    int a = (x - y);
    printf("a is %d\n", a);    // a is -1
    return 0;
}

_{(Live example: http://ideone.com/C3SlIn)}

What you're seeing there is the result of the integer promotions that are performed on the operands to - as part of the usual arithmetic conversions.¹ Relevant parts of the definition of integer promotions from the (C99) standard:

[6.3.1.1] If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int.

An int can indeed represent all values of a uint8_t, so that subtraction is really equivalent to:

int a = (int)x - (int)y;

In other words, there is no overflow.

---

_{1. But to preempt one common confusion, this behaviour is inherent to how - works; it's not because we're assigning to an int here.}