Call this function!

[LB]

template<typename T, typename U>
class e
{
    void Print() //call this function!
    {
        std::cout << "You haxxor..." << std::endl;
    }
public:
    T g()
    {
        return&U::Print;
    }
};

You cannot modify this code in any way, nor can to create preprocessor definitions that will modify the class. The point is to call the Print function from this class, the output is unimportant. You may create one class, and the class cannot extend the above class. You cannot do something hacky like casting function pointers.

Tip: If you can get an instance of the class, you're on the right track.

EDIT: My compiler is too stupid to know this isn't possible.

[hamsterman]

I think I've seen this here before. The problem is that e<A, B> does not have access to the privates of e<A, C>, so U would have to be an infinite type, which I don't think can be done.

Maybe this is different across compilers. What are you using?

[Catfish]

You may create one class, and the class cannot extend the above class. You cannot do something hacky like casting function pointers.

Cool, that. Is there anything else we cannot do?

Tip: If you can get an instance of the class, you're on the right track.

That's easy. To whomever wants to build further, here it is:
e<void (e<int, int>::*)(), e<int, int> > thing;

The problem is, thing.g() will still return the address of a private member function, and the GNU compiler will catch that.

Edit: layout.

[roberts]

e<void (e<int, int>::*)(), e<int, int> > thing;

Jeezus, I don't even grasp how that does what it's supposed to do. I'll never truly know how to program in C++.

[ne555]

¿what's the point of having 2 template parameters, if the second depends on the first if they are dependant?

Edit: To make it clear

template<class T>
class e{
  typedef typename T::print_type print_type;
public:
  print_type g();
};

[zwA4jE8b]

@catfish
What does the '*' do after the scope symbol? Make it a pointer?

[LB]

@Catfish: close, but you need the additional class I mentioned. Your code doesn't exactly compile on some compilers because e<int, int> has int g(){ return&int::Print; } (unless it is standard defined that unused functions need not be syntactically correct?)

[ne555]

@CreativeMFS: it's a pointer to a method.

typedef void (foo::*method)();

method m = &foo::bar; 
foo instance;
(instance.*m)(); //calls instance.bar() 

[hamsterman]

Here: http://ideone.com/Mki6P
The challenge is broken, accept it.

[LB]

Meh, guess my compiler is too stupid - just tested with my working code and it didn't work. Thanks, challenge closed.

[Catfish]

@ LB: Is that it? Don't disappoint me with a sterile retreat.
Where's your "working code", and which was the compiler for which it worked?

@ roberts: This isn't "programming", this is hackery, partly illustrating how horrible a language C++ is.

[LB]

Catfish it was exactly like yours but with Print being a function in the other class rather than a variable. I'll copypaste it when I get home to the computer that has it. The compiler is whichever one VC++2008 uses.