How to use ‘auto’ for declaring a pointer to an overloaded member function?

Question

289.76K viewsSeptember 10, 2022auto C# member-function-pointers

1

Itachi Uchiwa September 7, 2022 0 Comments

If I have an overloaded non-static member function, how can I use auto to declare a pointer for one version of that function?

struct Foo{
   void bar(){}
   void bar(int){}
};

auto ptr = &Foo::bar; // error: unable to deduce 'auto' from '&Foo::bar'

2 Answers

score 5 · Answer 1 · 2022-09-08T03:29:38+00:00

You don’t.

Well, you could do an explicit cast to the proper pointer type, but it’s really no different from just putting the proper pointer type in place of auto:

using bar_ptr = void (Foo::*)();

bar_ptr ptr = &Foo::bar;

//or

auto ptr = (bar_ptr)&Foo::bar;

score 2 · Answer 2 · 2022-09-08T02:06:51+00:00

You can avoid explicitly mentioning the full type through such a set of functions:

template <class... Params, class R>
    auto ovl(R(*f)(Params...)) {
    return f;
}

template <class... Params, class R, class T>
    auto ovl(R(T::*f)(Params...)) {
    return f;
}

template <class... Params, class R, class T>
    auto ovl_const(R(T::*f)(Params...) const) {
    return f;
}

Usage is as follows, providing the argument types of the targeted function:

auto ptr = ovl<int>(&Foo::bar);

See it live on Wandbox

Overloads for varargs, volatile, ref-qualified, etc. member functions are left as an exercise for the reader.