是i = i ++真的是一个未定义的行为？ [英] is i=i++ truly a undefined behavior?

问题描述

可能的重复：

任何人都可以解释这些未定义的行为（i = i ++ + ++ i，i = i ++，etc…）

根据c ++标准，

i = 3;
i = i++;

会导致未定义的行为。

我们使用术语未定义的行为，如果它可以导致多于一个结果。但是这里， i 的最终值将是4，无论评价的顺序如何，所以这不应该被称为未指定的行为？

We use the term "undefined behavior" if it can lead to more then one result. But here, the final value of i will be 4 no matter what the order of evaluation, so shouldn't this really be called "unspecified behavior"?

推荐答案

短语... i 的最终值将是4，评估顺序...不正确。编译器可以发出相当于下面的代码：

The phrase, "…the final value of i will be 4 no matter what the order of evaluation…" is incorrect. The compiler could emit the equivalent of this:

i = 3;
int tmp = i;
++i;
i = tmp;

或：

i = 3;
++i;
i = i - 1;

或此：

i = 3;
i = i;
++i;

对于术语的定义，如果答案被保证为4，

As to the definitions of terms, if the answer was guaranteed to be 4, that wouldn't be unspecified or undefined behavior, it would be defined behavior.

因为它是未定义的行为，根据标准（ Wikipedia ），因此它甚至可以这样做：

As it stands, it is undefined behaviour according to the standard (Wikipedia), so it's even free to do this:

i = 3;
system("sudo rm -rf /"); // DO NOT TRY THIS AT HOME … OR AT WORK … OR ANYWHERE.

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