i = i ++的行为真的不确定吗? [英] Is the behaviour of i = i++ really undefined?

问题描述

可能重复:
谁能解释这些未定义的行为(i = i ++ + ++ i，i = i ++，等等…)

Possible Duplicate:
Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)

根据c ++标准，

i = 3;
i = i++;

将导致不确定的行为.

我们使用术语未定义的行为"，如果它可以导致一个以上的结果.但是在这里，无论计算顺序如何，i的最终值都将为4，所以这真的不应该被称为未指定行为"吗?

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"...无论计算顺序如何，i的最终值将为4".编译器可以发出与之等效的东西:

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;

或者这个:

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

或者这个:

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

关于术语的定义，如果保证答案为4，则不会是未指定或未定义的行为，则将是已定义的行为.

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.

就目前而言，根据标准(维基百科)，它是未定义的行为，因此甚至可以免费这样做:

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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