*&++++ i是否会在C ++ 03中导致未定义的行为? [英] Does *&++i cause undefined behaviour in C++03?

问题描述

在另一个答案中指出，在C ++ 11之前，其中i是int ，然后使用表达式:

In another answer it was stated that prior to C++11, where i is an int, then use of the expression:

*&++i

引起不确定的行为.这是真的吗?

caused undefined behaviour. Is this true?

在另一个答案上，在评论中进行了一些讨论，但似乎没有说服力.

On the other answer there was a little discussion in comments but it seems unconvincing.

推荐答案

询问*&++i本身是否具有UB毫无意义.延迟不一定访问i的存储值(优先级或新值)，如将其用作引用的初始化表达式所见.只有涉及右值转换(在这种情况下使用)时，才有任何问题需要讨论.然后，由于我们可以使用++i的值，因此可以使用*&++i的值，但要与++i完全相同.

It makes little sense to ask whether *&++i in itself has UB. The deferencing doesn't necessarily access the stored value (prior or new) of i, as you can see by using this as an initializer expression for a reference. Only if an rvalue conversion is involved (usage in such context) is there any question to discuss at all. And then, since we can use the value of ++i, we can use the value of *&++i with exactly the same caveats as for ++i.

最初的问题本质上与i = ++i有关，与i = *&++i相同.在C ++ 03中，这是未定义的行为，这是因为i在序列点之间进行了两次修改，而在C ++ 11中则是定义良好的，这是由于赋值运算符的副作用在计算值之后进行了排序左右两侧.

The original question concerned essentially i = ++i, which is the same as i = *&++i. That was undefined behavior in C++03, due to i being modified twice between sequence points, and is well-defined in C++11, due to the side-effects of the assignment operator being sequenced after the value computations of the left and right hand sides.

可能需要注意的是，C ++ 98和C ++ 03标准中的非规范性示例是不正确的，将某些形式上未定义行为的情况描述为仅是未指定的行为.因此，意图一直没有完全清楚.一个好的经验法则是，不要仅仅依靠这种晦涩难懂的语言案例，而避免它们:避免为了理解代码而不必成为语言律师.

It is perhaps relevant to note that the non-normative examples in the C++98 and C++03 standards, were incorrect, describing some cases of formally Undefined Behavior as merely unspecified behavior. Thus, the intent has not been entirely clear, all the way back. A good rule of thumb is to simply not rely on such obscure corner cases of the language, to avoid them: one should not need to be a language lawyer in order to make sense of the code…

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