为什么x-> 0`不是未定义的行为，而`x = x--`是? [英] Why is `x-- > 0` not undefined behaviour, while `x = x--` is?

问题描述

众所周知，这会循环到零:

As everyone knows, this loops through zero:

while (x-- > 0) { /* also known as x --> 0 */
  printf("x = %d\n", x);
}

但是x = x--会产生不确定的行为.

两个示例都需要x--的一些返回"值，我猜这是不存在的.

Both examples need some 'return' value of x--, which is not there I guess. How can it be that x-- > 0 is defined but x = x-- is not?

推荐答案

因为在x = x--中，您两次修改了x的值，而没有中间的顺序点.因此，操作顺序未定义.在x-- > 0中，对x的值进行了一次修改，并且明确定义了x--的求值结果将是递减之前的x的值.

Because in x = x-- you're modifying the value of x twice without an intervening sequence point. So the order of operations is not defined. In x-- > 0 the value of x is modified once, and it is clearly defined that result of evaluating x-- will be the value of x before the decrement.

这篇关于为什么x-> 0`不是未定义的行为，而`x = x--`是?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！