问题描述
比方说，我有一个变量 i 来自外部资源：
Let's say I have a variable i that comes from external sources:
int i = get_i();
假设 i 为 INT_MIN 和二进制补码表示法， -i 是否未定义？
Assuming i is INT_MIN and two's complement representation, is -i undefined?
推荐答案
这取决于平台。 C支持负数的三种表示形式（请参见节）。如果是陷阱表示，则 -INT_MIN 等于 INT_MAX 。
With one's complement and sign and magnitude, -INT_MIN is defined (and equal to INT_MAX). With two's complement, it depends on whether the value with sign bit 1 and all value bits zero is a trap representation or a normal value. If it's a normal value, -INT_MIN overflows, resulting in undefined behavior (see section 6.5 of the C99 standard). If it's a trap representation, -INT_MIN equals INT_MAX.
也就是说，大多数现代平台使用的二进制补码没有陷阱表示，因此 -INT_MIN 通常会导致不确定的行为。
That said, most modern platforms use two's complement without trap representations, so -INT_MIN typically results in undefined behavior.
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