Why the member function address is the same

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Hi~ I am a student with very little exprience with C++.
Both printf() give me the same address when I run the following code (Compiled with Visual Studio 2010). I just can't understand why different member function can have the same address. Did I make any mistake?

C++

#include "stdafx.h"

class Base1
{
public:
	virtual void f() { }
};

class Base2
{
public:
	virtual void g() { }
};

class Sub : public Base1, public Base2
{
public:
	virtual void f() { cout<<"I am f()"<<endl; }
	virtual void g() { cout<<"I am g()"<<endl; }
};

int _tmain(int argc, _TCHAR* argv[])
{
	printf("0x%x\n",&Sub::f);
	printf("0x%x\n",&Sub::g);
        system("pause");
	return 0;
}

Posted 25-Oct-11 18:11pm

UMayLaugh

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Andrew Brock · Accepted Answer · 2011-10-25T19:05:00

The issue here is that they are virtual. This means that the function addresses are stored in a virtual table on a per-instance basis.

It is possible to get this information out of the per-instance vtable, however this is very unstable. I am not sure why the compiler wont give you direct access to this address.

Because you are also inheriting from 2 virtual classes, you should also be using virtual inheritance. class Sub : public virtual Base1, public virtual Base2

This is an explanation of how virtual functions are called, if you don't know:

Virtual functions are called in a different way to a normal function. If you go to the disassembly window and then go to the address it identifies as both functions you can see this.

Output:

0x4100a
0x4100a

Disassembly window:

Sub::`vcall'{0}':
0004100A  jmp         Sub::`vcall'{0}' (41120h)
[heaps of other stuff removed]
Sub::`vcall'{0}':
00041120  mov         eax,dword ptr [ecx]  
00041122  jmp         dword ptr [eax]

So what does this mean? it means that the same function is called every time you call a virtual function, and the address of the virtual function is stored at the memory address passed in as eax.

Think of it as the compiler changing this:

C++

Sub s;
s.f();
s.g();

With this:

C++

void CallVirtualFunc(void (*func)()) {
	func();
}

Sub s;
//Imagine we get the correct address for the following because they are'nt virtual functions
CallVirtualFunction(&Sub::f);
CallVirtualFunction(&Sub::g);

Then what you are getting is the address of the function CallVirtualFunc