Why must I use address-of operator to get a pointer to a member function?

Question

struct A
{
    void f() {}
};

void f() {}

int main()
{
    auto p1 = &f;     // ok
    auto p2 = f;        // ok
    auto p3 = &A::f; // ok

    //
    // error : call to non-static member function
    // without an object argument
    //
    auto p4 = A::f; // Why not ok?
}

What are you trying to do here? Why would you need the address of a member function, and if you did have such a thing, what would you do with it? — tadman, Feb 10 '17 at 01:46
You're playing with fire here. Why not use a [C++ lambda](http://en.cppreference.com/w/cpp/language/lambda) or a `static` wrapper function? — tadman, Feb 10 '17 at 01:49
@PaulRooney Is a member function going to be happy if severed from any instance? — tadman, Feb 10 '17 at 01:50
@tadman you have to pass the instance as well http://stackoverflow.com/questions/10673585/start-thread-with-member-function. — Paul Rooney, Feb 10 '17 at 01:51

WhiZTiM · Accepted Answer · 2017-02-10T02:01:38.867

auto p1 = &f;     // ok
auto p2 = f;      // ok

The first is more or less the right thing. But because non-member functions have implicit conversions to pointers, the & isn't necessary. C++ makes that conversion, same applies to static member functions.

To quote from cppreference:

An lvalue of function type T can be implicitly converted to a prvalue pointer to that function. This does not apply to non-static member functions because lvalues that refer to non-static member functions do not exist.

Why must I use address-of operator to get a pointer to a member function?

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