c++ - INT_MIN/-1 是 C++ 中定义的行为吗？

Question

我有以下 INT_MIN/-1 代码。我希望这会变成 INT_MAX+1(或翻转时为 0)。然而，我得到的实际结果是 INT_MIN。这是我的测试代码:

#define __STDC_LIMIT_MACROS
#include <stdint.h>
#include <stdio.h>
#include <limits.h>
using namespace std;
int main()
{
  int min=INT_MIN;
  int res=min/-1;
  printf("result: %i\n", res);
  printf("max: %i min: %i\n", INT_MAX, INT_MIN);
  return 0;
}

这个实现是特定的和/或未定义的行为吗？

Answer 1

Is this implementation specific and/or undefined behavior?

是的，signed 整数溢出是未定义的行为。根据 C++11 标准的第 5/4 段:

If during the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined. [...]

注意，这同样不适用于无符号算术。如第 3.9.1/4 段和脚注 46 中所述:

Unsigned integers, declared unsigned, shall obey the laws of arithmetic modulo 2^n where n is the number of bits in the value representation of that particular size of integer. [...]

This implies that unsigned arithmetic does not 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 unsigned integer type.