c++ 只有 : unary minus for 0x80000000

Question

这个问题应该是给语言律师的。

假设 signed 和 unsigned int 都是 32 位宽。如 n3337.pdf 草案 5.3.1.8 所述，

(-(0x80000000u)) = 0x100000000u-0x80000000u = 0x80000000u

但我找不到问题的答案:signed 0x80000000 的一元减号是什么？是 UB、实现定义还是...？

问题主要是关于运行时计算。

说

   signed int my_minus(signed int i) { return -i;}
   ....
   int main() {
       signed int a = -0x7FFFFFFF; // a looks like 0x80000001
       signed int b = a - 1;       // b looks like 0x80000000
       std::cout << my_minus(b);
       ....
   }

不过，欢迎您对其他 2 个案例发表评论:

• 编译时常量折叠，比如 -(INT_MIN)

• constexpr 的编译时计算(如果与编译时常量折叠有区别)。

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(请在重复投票前查看https://meta.stackexchange.com/questions/123713/is-splitting-a-question-a-good-practice。)

Answer 1

据我所知，有符号整数溢出总是未定义的。来自 C++ 规范部分 5 Expressions，第 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. [Note: most existing implementations of C++ ignore integer overflows. Treatment of division by zero, forming a remainder using a zero divisor, and all floating point exceptions vary among machines, and is usually adjustable by a library function. —endnote]