C / C++ / MFC

George_George wrote:

I think you mean even if the function has body in _interface, it is still treated as pure virtual?

George_George wrote:

How to prove it (is pure virtual)?

George_George wrote:

even if the function has body in _interface, it is still treated as pure virtual

George_George wrote:

I think even if you put the body into a pure virtual function, you can not utilize the function (body), right (for example. invoking it)? During compiling process, it is just discarded?

// VanillaConsole.cpp : Defines the entry point for the console application.
//
 
#include "stdafx.h"
#include <iostream>
using namespace std;
 
class Base
{
public:
	virtual void Func() = 0
	{
		cout << "Base::Func()" << endl;
	}
};
 
class Derived : public Base
{
public:
	virtual void Func()
	{
		cout << "Derived::Func()" << endl;
		Base::Func();
	}
};
 
int main(int argc, char* argv[])
{
	// Base b;
	//  If the above line is uncommented you'll get a compiler error.
 
	Derived d;
	d.Func();
 
	return 0;
}

George_George wrote:

In your code, you are invoking Func from class Derived. What I mean is you can not utilize Func from class Base. Right?

// VanillaConsole.cpp : Defines the entry point for the console application.
//
 
#include "stdafx.h"
#include <iostream>
using namespace std;
 
class Base
{
public:
	void CallItForMe()
	{
		Base::Func();
	}
 
	virtual void Func() = 0
	{
		cout << "Base::Func()" << endl;
	}
};
 
class Derived : public Base
{
public:
	virtual void Func()
	{
		cout << "Derived::Func()" << endl;
	}
};
 
int main(int argc, char* argv[])
{
	// Base b;
	//  If the above line is uncommented you'll get a compiler error.
 
	Derived d;
	d.CallItForMe();
 
	return 0;
}

George_George wrote:

Your code is wrong. I am discussing __interface, and if you change your code from class Base to __interface Base, your code can not compile.

// VanillaConsole.cpp : Defines the entry point for the console application.
//
 
#include "stdafx.h"
#include <iostream>
using namespace std;
 
class Base
{
public:
	virtual void Func() = 0
	{
		cout << "Base::Func()" << endl;
	}
};
 
class Derived : public Base
{
public:
	virtual void Func()
	{
		cout << "Derived::Func()" << endl;
	}
};
 
int main(int argc, char* argv[])
{
	// Base b;
	//  If the above line is uncommented you'll get a compiler error.
 
	Derived d;
	d.Base::Func();
 
	return 0;
}

// VanillaConsole.cpp : Defines the entry point for the console application.
//
 
#include "stdafx.h"
#include <iostream>
using namespace std;
 
__interface Base
{
	void Func()
	{
		cout << "Base::Func()" << endl;
	}
};
 
class Derived : public Base
{
public:
	virtual void Func()
	{
		cout << "Derived::Func()" << endl;
	}
};
 
int main(int argc, char* argv[])
{
	Derived d;
	d.Base::Func();
 
	return 0;
}

George_George wrote:

I have got your point, which is, in __interface we can put function body and implement and non-pure virtual function, and the function body is not ignored by compiler. Right?

George_George wrote:

"The pure-virtualness of a function and whether or not it contains a body are independent issues."

Why do you say it is independent issue?

__interface IFoo {

	int goo() {return 0;}
};

int main()
{
	cout << IFoo::goo() << endl;

	return 0;
}

__interface Base
{
public:	
virtual void Func() = 0	{		cout << "Base::Func()" << endl;	}
}; 

class Derived : public Base
{
public:	virtual void Func()	{		cout << "Derived::Func()" << endl;	}
}; 

int main(int argc, char* argv[])
{	
	Derived d;	
	d.Base::Func(); 	
	return 0;
}