[LordN3mrod]

Hi,
there was a time when I was sure that a typedef is simply meant for programmer's convenience, and that any program could theoretically be written without using it.
For example, I came to learn how to declare variables and/or functions with arbitrarily complicated types without a typedef, e.g. a function taking a pointer to function taking a reference to an array of 3 pointers to functions .... blah blah blah. I can do this without a typedef. But! I came across something which I cannot do without a typedef. Let's consider a functor object that doesn't have an operator()

struct Square
{
   static int f(int n) {return n*n;}
   typedef int(*pf)(int);
   operator pf () (return &f;)
}; 
#include <iostream>
int main()
{
   Square s;
   std::cout << s(4); //Outputs 16
}

Now I REALLY REALLY wanna know if there is a way to rewrite operator pf without a typedef, like operator int(*)(int) () {...} (of course it doesn't work this was).

Any help will be appreciated

[xpi0t0s]

Hmm very interesting! I wonder why this doesn't compile:

struct Square
{
   static int f(int n)
   {
	   return n*n;
   }
   static int f2(int n)
   {
	   return n+n;
   }
   typedef int(*pf)(int);
   typedef int(*pf2)(int);
   operator pf () 
   {
	   return &f;
   }
   operator pf2 () 
   { // <-- this line throws C2535
	   return &f2;
   }
};

Errors:

error C2535: 'Square::operator Square::pf(void)' : member function already defined or declared
see declaration of 'Square::operator Square::pf'

Edit: added code block to errors to prevent stupid smileys

[LordN3mrod]

why that code doesn't compile is obvious. You have defined the same member function twice.
pf and pf2 ARE NOT DISTICT TYPES!!! Therefore the two conversion operators are the same.

[xpi0t0s]

Ok smartass what about the OP's question.

[LordN3mrod]

whose? You mean my question ?

[xpi0t0s]

Note to self: pay more attention :-)