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

问题

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?
}

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

回答1:

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.

来源：https://stackoverflow.com/questions/42150125/why-must-i-use-address-of-operator-to-get-a-pointer-to-a-member-function