Thread: Address of a function

You've been told these things before but...
1) "#include <iostream>" along with "using namespace std;"
2) "int main()" and "return 0;" at the end.

About your problem, I think you should use:

Code:

p = e.display;

It says
line (24) : " '=' : cannot convert from 'void (__thiscall ex::*)(void)' to 'void (__cdecl *)(void)'"

Code:

#include<iostream>
class ex
{
	int i,j;
public: 

	ex(int ii,int jj)
	{
		i=ii;j=jj;
	}
	ex()
	{
		i=0;
	}
	void display()
	{
		std::cout<<i;
	}
};

int main()
{
	ex e(1,2); void (*p)();
	p=e.display;
	std::cout<<p;
return 0;
}

Then declare p as:

Code:

void (ex::*p)();

( I've never tried it myself, so don't start chasing me if it fails. )

Pointers to non-static member functions of a class are different types from pointers to other functions.

Try explaining what you are actually trying to achieve. Once you have "collected" the address of a function, what do you expect to be able to do with it?

A pointer to a member is not (necessarily) an address. The precise representation is implementation-specific, but is typically an offset relative to an object, not an address. That is why the syntax to call a pointer to member requires both an object as well as the pointer-to-member-function.

That is why I was asking you what you REALLY sought to achieve. Your question was meaningless. Your clarification still is.

And, if as grumpy says, you can't?