我对我写的这个模板很困惑，它应该“自动”推断出我​​传入的指针类型，但编译器似乎并不认为这两种类型是相同的，而我认为是，并且typeid() 在我这边。

#include <iostream>

template <auto* static_ptr_to_object>
struct Handle
{
    // static_ptr_to_object should be a pointer to a Whale<char>
    // Both of the following typedefs should result in types of Whale<char>
    // When I cout typeid() of these types it prints out the same types
    // However std::is_same_v returns false

    using pointee_type1 = std::remove_pointer_t<decltype(static_ptr_to_object)>;
    using pointee_type2 = decltype(*static_ptr_to_object); // The two are the same types

    void showTypes()
    {
        //static_assert(std::is_same_v<pointee_type1, pointee_type2>);
        // Uncommenting this line will make the static assert fail

        std::cout << "Type of 'pointee_type1' = " << typeid(pointee_type1).name() << "\n";
        std::cout << "Type of 'pointee_type2' = " << typeid(pointee_type2).name() << "\n";

        if (typeid(pointee_type1) == typeid(pointee_type2))
            std::cout << "Types are the same\n";
        else std::cout << "Types are not the same\n";

    }

    bool b1 = std::is_integral_v<decltype(pointee_type1::member)>;
    // Uncommenting the below line will make compilation fail

    //bool b2 = std::is_integral_v<decltype(pointee_type2::member)>;
    // pointee_type2 must be a class or namespace when followed by ::
    // pointee_type2 left of '::' must be a class or namespace name
    

};

template <typename T>
struct Whale
{
    T member;
};

Whale<char> whale;

int main()
{
    Handle<&whale> handleToWhale;
    handleToWhale.showTypes();
}

所以我认为这两种类型是相同的，并且 typeid() 运算符说它们是相同的，但是 std::is_same 返回 false，并且根据编译器错误消息，它无法识别 pointee_type2 是类类型，因为它无法识别::运算符。

编辑:我对解引用指针的理解是错误的，this question relates to在某些情况下，指针取消引用将如何返回引用类型。