How does the template parameter of std::function work? (implementation)

Question

In Bjarne Stroustrup\'s home page (C++11 FAQ):

struct X { int foo(int); };

std::function f;
f = &X::foo; //pointer to membe         


        

Answer 1

g++ seems to have an union which may keep either function pointer, member pointer or void pointer which probably points to a functor. Add overloads which appropriately flag which union member is valid and heavy casting to a soup and then it works...