c++ - 临时 const 数组未绑定(bind)到右值引用

Question

我有以下测试程序:

#include <iostream>
#include <type_traits>
#include <utility>

template<typename Ty, std::size_t N>
void foo(Ty (&&)[N])
{
    std::cout << "Ty (&&)[" << N << "]\t" << std::is_const<Ty>::value << '\n';
}

template<typename Ty, std::size_t N>
void foo(Ty (&)[N])
{
    std::cout << "Ty (&)[" << N << "]\t" << std::is_const<Ty>::value << '\n';
}

template<typename Ty>
using id = Ty;

int main()
{
    std::cout.setf(std::cout.boolalpha);

    foo(id<int[]>{1, 2, 3, 4, 5});
    foo(id<int const[]>{1, 2, 3, 4, 5}); // <-- HERE.
    int xs[]{1, 2, 3, 4, 5};
    foo(xs);
    int const ys[]{1, 2, 3, 4, 5};
    foo(ys);
    foo(std::move(xs));
    foo(std::move(ys));
}

我希望标有箭头的行会调用右值重载，就像它上面的非常量调用一样，但事实并非如此。

这只是 GCC 中的一个错误，还是标准中有什么东西导致选择左值重载？

Answer 1

根据标准§12.2 [class.temporary] :

Temporaries of class type are created in various contexts: binding a reference to a prvalue (8.5.3), returning a prvalue (6.6.3), a conversion that creates a prvalue (4.1, 5.2.9, 5.2.11, 5.4), throwing an exception (15.1), entering a handler (15.3), and in some initializations (8.5).

所以 id<int const[]>{1, 2, 3, 4, 5}是临时的，因此是纯右值 §3.10 [basic.lval] :

An rvalue (so called, historically, because rvalues could appear on the right-hand side of an assignment expression) is an xvalue, a temporary object (12.2) or subobject thereof, or a value that is not associated with an object.

A prvalue (“pure” rvalue) is an rvalue that is not an xvalue.

因此应选择具有右值引用参数的重载函数。