Why is a modulo operation returning an unexpected value

Question

Why is the following code printing 255?

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

int main(void) {
  uint8_t i = 0;
  i = (i - 1) % 16;
  printf("i: %d\n", i);
  return 0;
}

I assumed 15, although i - 1 evaluates to an integer.

Answer 1

Because of integer promotions in the C standard. Briefly: any type "smaller" than int is converted to int before usage. You cannot avoid this in general.

So what goes on: i is promoted to int. The expression is evaluated as int (the constants you use are int, too). The modulus is -1. This is then converted to uint8_t: 255 by the assignment.

For printf then i is integer-promoted to int (again): (int)255. However, this does no harm.

Note that in C89, for a < 0, a % b is not necessarily negative. It was implementation-defined and could have been 15. However, since C99, -1 % 16 is guaranteed to be -1 as the division has to yield the algebraic quotient.

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If you want to make sure the modulus gives a positive result, you have to evaluate the whole expression unsigned by casting i:

i = ((unsigned)i - 1) % 16;

Recommendation: Enable compiler warnings. At least the conversion for the assignment should give a truncation warning.