具有通用引用的成员函数模板将不接受左值

Question

I've been trying to use a template member function to set a value inside of my class. I wanted to use a universal reference so that I could accept any variant of the correct type (eg T , T& , T&& , const T , const T& , const T&& )

However, it seems that my member function will only accept rvalues, unlike a free function accepting a universal reference.

template <typename T>
class Foo{
public:
    void memberURef(T&& t){
        val = std::forward<T>(t);
    }
private:
    T val;
};


template <typename T>
void freeURef(T&& t){

}

int main() {
    int lval = 1;
    const int clval = 1;
    freeURef(lval); // fine
    freeURef(clval); // fine

    Foo<int> foo;
    foo.memberURef(2);
    foo.memberURef(lval); //error: cannot bind 'int' lvalue to 'int&&'
    foo.memberURef(clval); //error: no matching function for call to 'Foo<int>::memberURef(const int&)'
    return 0;
}

Answer 1

In the code given, Foo is instantiated as Foo<int> . Once this happens, the class template is instantiated as shown:

class Foo{
public:
    void memberURef(int&& t){       // Uh oh!
        val = std::forward<int>(t);
    }
private:
    int val;
};

Notice how the member function is no longer a template, and therefore no longer accepts a universal reference, but an rvalue reference. In order to make a member function that accepts a universal reference, the Foo class would need to be modified as follows:

template <typename T>
class Foo{
public:
    template <typename L>
    void memberURef(L&& t){
        val = std::forward<L>(t);
    }
private:
    T val;
};

That way, memberURef is still a function template after the class template has been instantiated, and it thus still accepts a universal reference.