是“++ l * = m”未定义的行为？ [英] Is "++l *= m" undefined behaviour?

问题描述

我开始学习C ++ 0x。我遇到了下面的表达式：

I have started studying about C++0x. I came across the follow expression somewhere:

int l = 1, m=2;
++l *= m;

我不知道第二个表达式是否具有良好定义的行为。所以我在这里问。

I have no idea whether the second expression has well defined behavior or not. So I am asking it here.

不是UB吗？

推荐答案

在上面的代码中，前缀 ++ 优先于 * = ，因此首先执行。结果是 l 等于 4 。

In the code above, prefix ++ has precedence over *=, and so gets executed first. The result is that l equals 4.

UPDATE：确实是未定义的行为。

UPDATE: It is indeed undefined behavior. My assumption that precedence ruled was false.

原因是 l 是<$ p $的左值和右值c $ c> * = ，以及 ++ 。这两个操作不排序。因此， l 被写入（和读取）两次没有序列点（旧标准字），行为是未定义的。

The reason is that l is both an lvalue and rvalue in *=, and also in ++. These two operations are not sequenced. Hence l is written (and read) twice "without a sequence point" (old standard wording), and behavior is undefined.

作为旁注，我推测你的问题源于C ++ 0x中序列点的变化。 C ++ 0x已经将关于序列点的措辞改为之前排序，以使标准更清晰。据我所知，这不改变C ++的行为。

As a sidenote, I presume your question stems from changes regarding sequence points in C++0x. C++0x has changed wording regarding "sequence points" to "sequenced before", to make the standard clearer. To my knowledge, this does not change the behavior of C++.

更新2：事实证明，按照 N3126草稿的C ++ 0x 。 @Johannes Schaub的回答是正确的，并记录语句的排序。信用当然应该回答他的回答。

UPDATE 2: It turns out there actually is a well defined sequencing as per sections 5.17(1), 5.17(7) and 5.3.2(1) of the N3126 draft for C++0x. @Johannes Schaub's answer is correct, and documents the sequencing of the statement. Credit should of course go to his answer.

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