Get address of const method

Question

I would like to be able to form a pointer-to-member type knowing only the class itself and method name. Unfortunately I am not able to do so having a const and non-const method variants in my class.

Example code snippet:

struct A
{
    void method() {}
    void method() const {}
};

int main()
{
    decltype(&A::method) _;
}

I have also tried the following this but without much success either:

decltype(&(std::declval<const A>().method)) _;

Both approaches fail as decltype cannot resolve this due to ambiguity:

'decltype cannot resolve address of overloaded function'

How can I achieve this in some other way?

Answer 1

You can do it like this:

struct A
{
    void method() {
        cout << "Non const\n";    
    }
    void method() const {
        cout << "const function\n";
    }
};

int main()
{
    typedef  void (A::*method_const)() const;
    method_const a = &A::method;     //address of const method
    
    typedef  void (A::*method_nonconst)();
    method_nonconst b = &A::method;  //address of non const method
    
    A var;
    std::invoke(a, var);
    std::invoke(b, var);
}

If you want to use decltype() to achieve the same thing, you first have to manually select the function you want using static_cast<>:

int main()
{
    //const
    decltype( static_cast <void (A::*)() const> (&A::method) ) a;
    //non const
    decltype( static_cast <void (A::*)()> (&A::method) ) b;
    
    a = &A::method;
    b = &A::method;
    
    A var;
    std::invoke(a, var);
    std::invoke(b, var);
}

Live on Coliru