初始化是否需要左值到右值的转换?是`int x = x;` UB吗?

免责声明:我承认这个答案相当投机.另一方面，当前 C++11 标准的表述似乎不允许更正式的答案.

在 这个问答，C++11标准未能正式指定值类别是每种语言结构所期望的.在下文中，我将主要关注 内置运算符，尽管问题是关于 initializers.最后，我会将我对运算符的情况得出的结论扩展到初始化程序的情况.

In the context of this Q&A, it has emerged that the C++11 Standard fails to formally specify what value categories are expected by each language construct. In the following I will mostly focus on built-in operators, although the question is about initializers. Eventually, I will end up extending the conclusions I drew for the case of operators to the case of initializers.

在内置运算符的情况下，尽管缺乏正式规范，但在标准中发现(非规范)证据表明意图规范是让prvalues 在任何需要值的地方都需要，如果没有另外指定.

In the case of built-in operators, in spite of the lack of a formal specification, (non-normative) evidences are found in the Standard that the intended specification is to let prvalues be expected wherever a value is needed, and when not specified otherwise.

例如，第 3.10/1 段中的注释说:

For instance, a note in Paragraph 3.10/1 says:

第 5 章中对每个内置运算符的讨论指出了它产生的值的类别以及它所期望的操作数的值类别.例如，内置赋值运算符期望左操作数是左值，右操作数是纯右值并产生左值作为结果.用户定义的运算符是函数，而类别他们期望的值和产量由他们的参数和返回类型决定

The discussion of each built-in operator in Clause 5 indicates the category of the value it yields and the value categories of the operands it expects. For example, the built-in assignment operators expect that the left operand is an lvalue and that the right operand is a prvalue and yield an lvalue as the result. User-defined operators are functions, and the categories of values they expect and yield are determined by their parameter and return types

另一方面，关于赋值运算符的第 5.17 节没有提到这一点.但是，再次在注释(第 5.17/1 段)中提到了执行左值到右值转换的可能性:

Section 5.17 on assignment operators, on the other hand, does not mention this. However, the possibility of performing an lvalue-to-rvalue conversion is mentioned, again in a note (Paragraph 5.17/1):

因此，函数调用不应干预左值到右值的转换和与任何单个复合赋值运算符相关的副作用

Therefore, a function call shall not intervene between the lvalue-to-rvalue conversion and the side effect associated with any single compound assignment operator

当然，如果不期望右值，则此注释将毫无意义.

Of course, if no rvalue were expected, this note would be meaningless.

正如 Johannes Schaub 在评论中指出的那样，在 4/8 中发现了另一个证据链接问答:

Another evidence is found in 4/8, as pointed out by Johannes Schaub in the comments to linked Q&A:

在某些情况下，某些转化会被抑制.例如，左值到右值的转换不是在一元 & 的操作数上进行的.操作员.在这些运算符和上下文的描述中给出了特定的例外情况.

There are some contexts where certain conversions are suppressed. For example, the lvalue-to-rvalue conversion is not done on the operand of the unary & operator. Specific exceptions are given in the descriptions of those operators and contexts.

这似乎暗示左值到右值的转换是在内置运算符的所有操作数上执行的，除非另有说明.反过来，这意味着 除非另有说明，否则期望右值作为内置运算符的操作数.

This seems to imply that lvalue-to-rvalue conversion is performed on all operands of built-in operators, except when specified otherwise. This would mean, in turn, that rvalues are expected as operands of built-in operators unless specified otherwise.

猜想:

尽管初始化不是赋值，因此运算符不参与讨论，但我怀疑规范的这个区域受到上述相同问题的影响.

Even though initialization is not assignment, and therefore operators do not enter the discussion, my suspicion is that this area of the specification is affected by the very same problem described above.

支持这一信念的痕迹甚至可以在第 8.5.2/5 段中找到，关于 references 的初始化(不需要左值初始化表达式的值):

Traces supporting this belief can be found even in Paragraph 8.5.2/5, about the initialization of references (for which the value of the lvalue initializer expression is not needed):

不需要通常的左值到右值(4.1)、数组到指针(4.2)和函数到指针(4.3)标准转换，因此被禁止，当这种对左值的直接绑定完成时.

The usual lvalue-to-rvalue (4.1), array-to-pointer (4.2), and function-to-pointer (4.3) standard conversions are not needed, and therefore are suppressed, when such direct bindings to lvalues are done.

通常"一词似乎暗示在初始化非引用类型的对象时，将应用左值到右值的转换.

The word "usual" seems to imply that when initializing objects which are not of a reference type, lvalue-to-rvalue conversion is meant to apply.

因此，我认为尽管对初始化器的期望值类别的要求没有明确规定(如果不是完全缺失的话)，但基于所提供的证据，假设 有意规范是:

Therefore, I believe that although requirements on the expected value category of initializers are ill-specified (if not completely missing), on the grounds of the evidences provided it makes sense to assume that the intended specification is that:

只要语言结构需要一个值，除非另有说明，否则需要一个纯右值.

在这种假设下，在您的示例中需要进行左值到右值的转换，这会导致未定义的行为.

Under this assumption, an lvalue-to-rvalue conversion would be required in your example, and that would lead to Undefined Behavior.

其他证据:

只是为了提供进一步的证据来支持这个猜想，让我们假设它错误，因此复制初始化确实不需要左值到右值的转换，并考虑以下代码(感谢jogojapan 贡献):

Just to provide further evidence to support this conjecture, let's assume it wrong, so that no lvalue-to-rvalue conversion is indeed required for copy-initialization, and consider the following code (thanks to jogojapan for contributing):

int y;
int x = y; // No UB
short t;
int u = t; // UB! (Do not like this non-uniformity, but could accept it)
int z;
z = x; // No UB (x is not uninitialized)
z = y; // UB! (Assuming assignment operators expect a prvalue, see above)
       // This would be very counterintuitive, since x == y

这种不统一的行为对我来说没有多大意义.IMO 更有意义的是，只要需要一个值，就需要一个纯右值.

This non-uniform behavior does not make a lot of sense to me. What makes more sense IMO is that wherever a value is required, a prvalue is expected.

此外，正如 Jesse Good 在他的回答中正确指出的那样，C++ 标准的关键段落是 8.5/16:

Moreover, as Jesse Good correctly points out in his answer, the key Paragraph of the C++ Standard is 8.5/16:

——否则，被初始化对象的初始值为(可能已转换)初始化表达式的值.标准如有必要，将使用转换(第 4 条)来转换初始化表达式为 cv 非限定版本目的地类型；不考虑用户定义的转换.如果无法进行转换，初始化格式错误.[ 笔记:cv1 T"类型的表达式可以初始化cv2 T"类型的对象独立于 cv 限定符 cv1 和 cv2.

— Otherwise, the initial value of the object being initialized is the (possibly converted) value of the initializer expression. Standard conversions (Clause 4) will be used, if necessary, to convert the initializer expression to the cv-unqualified version of the destination type; no user-defined conversions are considered. If the conversion cannot be done, the initialization is ill-formed. [ Note: An expression of type "cv1 T" can initialize an object of type "cv2 T" independently of the cv-qualifiers cv1 and cv2.

然而，虽然 Jesse 主要关注if necessary"这一点，但我还想强调type"这个词.上面的段落提到将使用标准转换如果需要"转换为目标类型，但没有说明类别转换:

However, while Jesse mainly focuses on the "if necessary" bit, I would also like to stress the word "type". The paragraph above mentions that standard conversions will be used "if necessary" to convert to the destination type, but does not say anything about category conversions:

1. 是否会根据需要进行类别转换?
2. 他们需要吗?

关于第二个问题，正如答案的原始部分所讨论的那样，C++11 标准目前没有指定是否需要类别转换，因为没有提到复制初始化是否需要 prvalue作为初始化器.因此，不可能给出明确的答案.但是，我相信我提供了足够的证据来假设这是预期规范，因此答案将是是".

For what concerns the second question, as discussed in the original part of the answer, the C++11 Standard currently does not specify whether category conversions are needed or not, because nowhere it is mentioned whether copy-initialization expects a prvalue as an initializer. Thus, a clear-cut answer is impossible to give. However, I believe I provided enough evidence to assume this to be the intended specification, so that the answer would be "Yes".

至于第一个问题，我认为答案也是是"是合理的.如果它是否"，那么显然正确的程序将是格式错误的:

As for the first question, it seems reasonable to me that the answer is "Yes" as well. If it were "No", obviously correct programs would be ill-formed:

int y = 0;
int x = y; // y is lvalue, prvalue expected (assuming the conjecture is correct)

总结一下(A1 = "第 1 题答案", A2 = "第 2 题答案"):

To sum it up (A1 = "Answer to question 1", A2 = "Answer to question 2"):

          | A2 = Yes   | A2 = No |
 ---------|------------|---------|
 A1 = Yes |     UB     |  No UB  | 
 A1 = No  | ill-formed |  No UB  |
 ---------------------------------

如果 A2 为否"，A1 无关紧要:没有 UB，但第一个例子的奇怪情况(例如 z = y 给出 UB，但没有 z = x 即使 x == y) 出现.另一方面，如果 A2 为是"，则 A1 变得至关重要；然而，已经有足够的证据证明它会是是".

If A2 is "No", A1 does not matter: there's no UB, but the bizarre situations of the first example (e.g. z = y giving UB, but not z = x even though x == y) show up. If A2 is "Yes", on the other hand, A1 becomes crucial; yet, enough evidence has been given to prove it would be "Yes".

因此，我的论点是 A1 = "Yes" 和 A2 = "Yes"，我们应该有未定义的行为.

Therefore, my thesis is that A1 = "Yes" and A2 = "Yes", and we should have Undefined Behavior.

进一步的证据: