标准 type_traits 库会产生一些使用 decltype 的类型别名的意外行为

Question

让我举例说明。我们以 std::is_same_v 和 std::is_base_of_v 为例。考虑以下代码：

#include <iostream>
#include <array>
#include <type_traits>

using namespace std;

struct base {};
struct derived : base { int foo; };

array<derived, 10> my_array;

int main()
{
    using c1 = decltype(*begin(my_array));
    using c2 = derived;

    if constexpr(is_same_v<c1,c2>)
    {
        cout<<"Correct!"<<endl;
    }
    else
    {
        cout << "No luck even though:" << endl
             << "c1 type is " << typeid(c1).name() << endl
             << "c2 type is " << typeid(c2).name() << endl;
    }

    if constexpr(is_base_of_v<base, c1>)
    {
        cout<<"Correct!"<<endl;
    }
    else
    {
        cout << "No luck even though:" << endl
             << "is_base_of_v<base, derived> = " << is_base_of_v<base, derived> << endl;
    }

    return 0;
}

预期结果是：

Correct!
Correct!

但是，在 clang 10 和 gcc 11 中的实际结果是：

No luck even though:
c1 type is 7derived
c2 type is 7derived
No luck even though:
is_base_of_v<base, derived> = 1

头脑发热。我主要感兴趣的是找出此行为背后的确切原因，然后可能会找到解决方法。如果可能的话，解决方法应该适用于任何可迭代类型，并且只适用于 std::array.

Answer 1

你的问题是你有额外的参考：

using c1 = decltype(*begin(my_array)); // derived&
using c2 = derived;

is_same_v<derived, derived&> 为假。

std::decay_t or std::remove_reference_t 可能会有帮助。

与typeid相同：

Refers to a std::type_info object representing the type type. If type is a reference type, the result refers to a std::type_info object representing the referenced type.

另一种了解类型的方法是使用该技巧：

template <typename> struct Debug; /*No definition*/

Debug<c1> d; // Incomplete type

错误消息类似于

error: aggregate 'Debug<derived&> d' has incomplete type and cannot be defined

Demo

标准 type_traits 库会产生一些使用 decltype 的类型别名的意外行为

The standard type_traits library produces unexpected beavior with some type aliases that use decltype

c++

c++17

typetraits