Question

Here is well described how to call member function by pointer:
http://www.newty.de/fpt/functor.html

But the functor needs to get 2 arguments: pointer-to-object and pointer-to-member-function:

TSpecificFunctor(TClass* _pt2Object, void(TClass::*_fpt)(const char*))
{ pt2Object = _pt2Object;  fpt=_fpt; }

call:

(*pt2Object.*fpt)(string);

Is it possible to pass single argument like C-style:

func() -- call
func -- function pointer

Why obj.method isn’t complete pointer-to-class-member?

Answer 1

The syntax object.*ptmf doesn’t create an intermediate object. It has no meaning and is forbidden by the language. You have to immediately call the result of accessing a pointer to member function.

You can explicitly create such an object using std::bind, which interprets the ptmf as a functor object, and makes the implicit this argument explicit.

auto fn = std::bind( ptmf, object, std::placeholders::_1 );
std::function< void( const char * ) > stdfn = fn;
fn( "foo" ); // equivalent to object.*ptmf( "foo" );

http://ideone.com/ds24F

Note that this functionality is new in C++11. Although C++03 TR1 has function and bind, they won’t perform this conversion on a ptmf. (Plain C++03 can do the job with std::mem_fn and std::bind1st, but they are extremely painful to use and have been deprecated.)