Question about C behaviour for unsigned integer underflow

Question

I have read in many places that integer overflow is well-defined in C unlike the signed counterpart.

Is underflow the same?

For example:

unsigned int x = -1; // Does x == UINT_MAX?

Thanks.

I can't recall where, but i read somewhere that arithmetic on unsigned integral types is modular, so if that were the case then -1 == UINT_MAX mod (UINT_MAX+1).

Answer 1

-1, when expressed as a 2's complement number, amounts to 0xFF...F for how ever many bits your number is. In an unsigned number space that value is the maximum value possible (i.e. all the bits are set). Therefore yes, x == UINT_MAX. The following code emits "1" on a C99 strict compiler:

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

int main(int argc, char **argv){
  uint32_t x = -1;      
  printf("%d", x == UINT_MAX ? 1 : 0);
  return 0;
}

Answer 2

You are mixing signed and unsigned numbers, which is uncool.

unsigned int x = 0u - 1u; // is OK though