c++ - 为什么 std::is_copy_constructible 的行为不如预期？

Question

#include <type_traits>

int main()
{
    std::is_constructible_v<int&, const int&>; // false, as expected.
    std::is_copy_constructible_v<int&>; // true, NOT as expected!
}

根据 cppref :

If T is an object or reference type and the variable definition T obj(std::declval()...); is well-formed, provides the member constant value equal to true. In all other cases, value is false.

std::is_copy_constructible_v<int&>应该给出与 std::is_constructible_v<int&, const int&> 相同的结果做;然而，clang 7.0给出不同的结果，如上所示。

这种行为是否符合 C++ 标准？

Answer 1

is_copy_constructible的引用是什么？状态是:

If T is not a referenceable type (i.e., possibly cv-qualified void or a function type with a cv-qualifier-seq or a ref-qualifier), provides a member constant value equal to false. Otherwise, provides a member constant value equal to std::is_constructible<T, const T&>::value.

所以，这里 is_copy_constructible<T>::value与std::is_constructible<T, const T&>::value相同.

所以在你的情况下:

std::is_constructible<int, const int&>::value将与 std::is_copy_constructible_v<int> 相同.

参见 DEMO