可以reinterpret_cast（或任何演员）将xvalues转换为lvalue？

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问题描述

以下代码是否合法（通过C ++ 11和/或C ++ 14标准）？

Is the following code legal (by C++11 and/or C++14 standard(s))?

#include <iostream>
#include <utility>

using namespace std;

void foo(int &a) {
    cout << a << endl;
}

int main() {
    foo(reinterpret_cast<int &>(move(5)));
}

如果是，是否未定义行为？ / li>
如果不是未定义的行为，我甚至可以在内它变成UB？

 
 
它在clang 3.5上编译，而不是在gcc 4.9上编译。 
 GCC错误：
It compiles on clang 3.5, not on gcc 4.9.GCC error:
➤ g++-4.9 -std=c++1y sample.cpp -o sample
sample.cpp: In function 'int main()':
sample.cpp:11:40: error: invalid cast of an rvalue expression of type 'std::remove_reference<int>::type {aka int}' to type 'int&'
     foo(reinterpret_cast<int &>(move(5)));
                                        ^
 
 
 
 EDIT 
 
 
  FYI，一个定制的铸造，比以前更少毛，它工作在C ++ 11为GCC和Clang，将是以下 lvalue 函数：
#include <iostream>

namespace non_std {
    template <typename T>
    constexpr T &lvalue(T &&r) noexcept { return r; }
}

void divs(int &a, int &b) {
    int t = a;
    a /= b;
    b /= t;
}

int main() {
    using namespace std;
    using namespace non_std;

    int i_care_for_this_one = 4;
    divs(i_care_for_this_one, lvalue(2));
    cout << i_care_for_this_one << endl;
}
 
 
推荐答案
  / strong>代码在C ++ 11中是不成形的。下面的答案是为C ++ 14。 
Update: The code is ill-formed in C++11. Answer below is for C++14. See note at the end of this answer.
我相信此代码格式正确和定义良好。这是为什么。
I believe this code is both well-formed and well-defined. Here's why.
  std :: move 的结果是xvalue [1] ;并将标量转换为一个左值引用 reinterpret_cast 似乎被标准的措辞允许：
The result of std::move is an xvalue [1], which is a type of glvalue; and converting a glvalue to an lvalue reference with reinterpret_cast appears to be allowed by the wording of the standard:
由于 int 的指针可以转换为指向 int   reinterpret_cast 也是允许的。该标准没有说明目标类型是否是一个左值引用或右值引用。
Since "pointer to int" can be converted to "pointer to int", this reinterpret_cast is also allowed. The standard doesn't say anything about whether the destination type has to be an lvalue reference or rvalue reference.
转换的结果是由上述段落：它引用与源glvalue相同的对象 - 即，值为 5的临时 int 对象。 （[dcl.init.ref]指定当一个prvalue绑定到一个引用时创建一个临时变量。）
The result of the cast is well-defined by the paragraph above: it refers to the same object as the source glvalue---that is, a temporary int object with the value 5. ([dcl.init.ref] specifies that a temporary is created when a prvalue is bound to a reference.)
通过 int& 也不违反任何别名规则，因为原始对象也是 int 类型。事实上，我相信，通过这样获得的左值来修改临时值也是很明确的。
Accessing the value through the int& also doesn't violate any aliasing rules since the original object was also of type int. In fact I believe it would even be well-defined to modify the temporary through the lvalue thus obtained.
 注意：措辞表示左值表达式，而不是glvalue表达式。与glvalue expression的措辞是从N3936，这是C ++ 14的最后工作草案。我不是标准化过程如何工作的专家，但我相信这意味着左值到glvalue的变化已经被委员会投票，当ISO发布C ++ 14标准时，它会非常类似于上面所说的。
Note: The C++11 wording says "lvalue expression", not "glvalue expression". The wording with "glvalue expression" is from N3936, which is the final working draft for C++14. I'm not an expert in how the standardization process works, but I believe this means that the change of "lvalue" to "glvalue" was already voted in by the committee, and when ISO publishes the C++14 standard, it's going to be pretty similar to what it says above.
 [1]除了在极少数情况下，参数是一个函数;在这种情况下，结果是一个左值，因为没有函数rvalues。
[1] Except in the rare case in which the argument is a function; in that case the result is an lvalue, since there are no function rvalues.
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