Access a private virtual member-function through pointers - c++

Question

class foo{
public:
int n;
private:
virtual void sayHi(){
cout<<"Hi there!";
}
};
How do I get the address of sayHi()??
main(){
foo f;
typedef void(*fptr)();
fptr func = reinterpret_cast<fptr>((&f)[0]);
(*func)();
}
The code above didn't work.
I know that the first 8 bytes of "f" object is a pointer to a virtual table where it contains the pointers to the functions, I'm using 64-bits machine. I'm basically trying to call the sayHi() through its pointer rather than calling it directly from f, since sayHi() is private anyway! How would I do this? Am I apportioning it right??

Answer 1

You cannot simply have a pointer to a non-static member function and call it without an object. The most straight-forward way to interpret your question would be using a pointer-to-member as in:
auto fptr = &foo::sayHi;
foo f;
(f.*fptr)();
Now, you say you want to call it without having to go through f. It's not clear exactly what this means. Using a lambda is probably good enough to create a callable that works as you want
auto func = [] { return foo{}.sayHi(); };
func(); // call
or using a specific object and capturing it by reference (shown) or by value
foo f;
auto func = [&f] { return f.sayHi(); };
func();

Answer 2

sayHi() is a non-static method of a class. You need to use a pointer-to-method instead of a raw pointer (the implementation of a method pointer is compiler-specific, so assuming the internal layout of the pointer-to-method is not portable).
Also, sayHi() is private to foo, so main() cannot access it directly. You need to either:
declare sayHi() as public:
class foo
{
int n;
public:
virtual void sayHi(){
cout << "Hi there!";
}
};
int main()
{
typedef void (foo::*fptr)();
fptr func = &foo::sayHi;
foo f;
(f.*func)();
return 0;
}
make main() be a friend of foo:
class foo
{
int n;
virtual void sayHi(){
cout << "Hi there!";
}
friend int main();
};
int main()
{
typedef void (foo::*fptr)();
fptr func = &foo::sayHi;
foo f;
(f.*func)();
return 0;
}

Answer 3

Well, ISO C++ forbids taking the address of a bound member function to form a pointer to member function. However, you can do something like this, if it helps. You may check the result here.
#include <iostream>
using namespace std;
class foo{
int n;
public:
virtual void sayHi(){
cout<<"Hi there!";
}
};
typedef void(*fptr)();
int main() {
auto func = reinterpret_cast<fptr>(&foo::sayHi);
(*func)();
return 0;
}