Programming Forums > Application Development > C++

Klarre · Mar 27th, 2007, 11:02 AM

Is there an easy way to get the memory address to a member function? I don't really trust (understand) the output of this code...

(Toggle Plain Text)

#include <iostream>

class Foo
{
public:
	void func()
	{
		std::cout << func << std::endl;
	}

	void func2()
	{
		std::cout << func2 << std::endl;
	}
};

int main()
{
	Foo foo;
	foo.func();
	foo.func2();

	return 0;
}

(Toggle Plain Text)

Output:
1
1

Why does it writes "1" to the screen in both functions? How do I make it print the real memory address?

Thanks for your help!

/Klarre

Game_Ender · Mar 27th, 2007, 9:48 PM

A member function doesn't have a unique address for each object, you need both a pointer to the member function an object to call it with. The the C++ FAQ.

Klarre · Mar 28th, 2007, 1:07 AM

Quote:

Originally Posted by Game_Ender

A member function doesn't have a unique address for each object.

Of course. But still the function have to be somewhere in the memory space, at a specific memory address. Guess I will need some assembly for this, though the "pointer-to-member" pointer works in a different manner than "pointer-to-function" does. Thanks for the link!

/Klarre

Eoin · Mar 28th, 2007, 6:34 AM

I believe you need to fully qualify the function name, i.e. &Foo::member_func. Here is a note from Microsoft on it-

Quote:

Pointer-to-members now require qualified name and &

Code written for previous versions of the compiler that just used the method name will now give Compiler Error C3867 or Compiler Warning C4867. This diagnostic is required by Standard C++. Now, to create a pointer to a member function, the address of operator (&) must be used with the fully qualified name of the method. Not having to use the & operator and the fully qualified name of the method can lead to logic bugs in code due missing parentheses in function calls. Using the function's name without an argument list results in a function pointer which is convertible to several types. This code would have compiled, leading to unexpected behavior at runtime.

Mar 27th, 2007, 11:02 AM	#1
Klarre Game engine designer Join Date: May 2005 Location: Sweden Posts: 297 Rep Power: 4	Get member function address Is there an easy way to get the memory address to a member function? I don't really trust (understand) the output of this code... (Toggle Plain Text) #include <iostream> class Foo { public: void func() { std::cout << func << std::endl; } void func2() { std::cout << func2 << std::endl; } }; int main() { Foo foo; foo.func(); foo.func2(); return 0; } #include <iostream> class Foo { public: void func() { std::cout << func << std::endl; } void func2() { std::cout << func2 << std::endl; } }; int main() { Foo foo; foo.func(); foo.func2(); return 0; } (Toggle Plain Text) Output: 1 1 Output: 1 1 Why does it writes "1" to the screen in both functions? How do I make it print the real memory address? Thanks for your help! /Klarre __________________ http://www.klarre.se

Mar 27th, 2007, 9:48 PM	#2
Game_Ender Professional Programmer Join Date: May 2006 Location: Maryland, USA Posts: 306 Rep Power: 3	A member function doesn't have a unique address for each object, you need both a pointer to the member function an object to call it with. The the C++ FAQ. __________________ Robotics @ Maryland AUV Team - Software Lead

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Programming Forums > Application Development > C++

Get member function address