Re: Figure out why member function pointer doesn't work?

On 5 Apr., 03:41, Nephi Immortal wrote:


First of all, you need to recognize that the precedence of operator->
is higher than of the operator->*:

  x->y->z is grouped like this: (x->y)->z

while

  x->*y->z is grouped like this: x->*(y->z)

Secondly, what you are trying to do cannot work without a proxy
callable object. In order to make

  (x->*y)();

work your operator->* function would need to return something that is
callable. Unfortunately, the result of ptr->*pmemfun where ptr is a
raw pointer and pmemfun is a member function pointer can only be used
as operand for the function call operator. You cannot return the
result of ptr->*pmemfun from a function and delay the function call
this way. C++ does not allow this.

If you want to support operator->* to be able to invoke member
functions you could try somehting like this:

  struct bound_data_memfun {
    data* ptr;
    pGo pmemfun;
    void operator()() const {return (ptr->*pmemfun)();}
    // ^^^^^^^^^^^^^
    // can only be used in combinaton with the function call
    // operator.
  };

  struct storage {
    ...
    bound_data_memfun operator->*(pGo pmemfun)
    {
      bound_data_memfun result = {pData,pmemfun};
      return result;
    }
    ...
  };

IMHO it's not worth the hassle, though. None of the popular smart
pointer classes overload the operator->* and there is no harm in that.
Simply write

  ((*x).*y)();

and be done.

SG