Question about C behaviour for unsigned integer underflow

问题

I have read in many places that unsigned 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).

回答1:

~~§6.2.5, paragraph 9:~~

A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type.

Edit:

Sorry, wrong reference, but the result is still pinned down. The correct reference is §6.3.1.3 (signed and unsigned integer conversion):

if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.

So yes, x == UINT_MAX.

回答2:

-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;
}

回答3:

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

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

来源：https://stackoverflow.com/questions/2760502/question-about-c-behaviour-for-unsigned-integer-underflow

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integer

types

integer-overflow

underflow