About sizeof of a class member function pointer [duplicate]

Question

Let's say we have a class A

class A;

and these typedefs

typedef void (A::*a_func_ptr)(void);
typedef void (*func_ptr)(void);

My question is why sizeof(a_func_ptr) returns 16, while sizeof(func_ptr) returns 4 (as for any pointer on x86 system)?

For instance

int main(int argc, char *argv[])
{
  int a = sizeof(a_func_ptr);
  int b = sizeof(func_ptr);
}

Answer 1

My question is why sizeof(a_func_ptr) returns 16, while sizeof(func_ptr) returns 4 (as for any pointer on x86 system)?

Because pointer-to-members are implemented differently. They're not pointers under the hood. Some compilers, such as MSVC, implement them as struct with more than one members in it.

Read this interesting article:

• Pointers to member functions are very strange animals

Note that in some compilers, they might have same size. The bottomline is: they're compiler-dependent.

Answer 2

Consider the following:

#include <iostream>

class A {
public:
    virtual void foo(){ std::cout << "A::foo" << std::endl; }
    void bar(){ std::cout << "A::bar" << std::endl; }
};

class B : public A {
public:
     void foo(){  std::cout << "B::foo" << std::endl; }
     void bar(){ std::cout << "B::bar" << std::endl; }
};

typedef void (A::*a_func_ptr)(void);

int main() {
   a_func_ptr f = &A::foo;
   a_func_ptr g = &A::bar;
   B b;
   A a;
   (b.*f)();
   (b.*g)();
   (a.*f)();
   (a.*g)();
}

Output:

B::foo
A::bar
A::foo
A::bar

Both member pointers are of the same type, yet both correctly routed the call in every cases.

Somehow, the generated programme must know when a pointer to method is actually a simple method or a virtual one. Thus the runtime representation of a method pointer has to include more information to handle the second case.

_{Remark: the size seems to be implementation dependent (I get 8 on my system).}