C / C++ / MFC

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__interface IFoo {

	virtual int foo() = 0;
	int goo() {return 0;}
};

// VanillaConsole.cpp : Defines the entry point for the console application.
//
 
#include "stdafx.h"
#include <iostream>
using namespace std;
 
// This class will compile just find - pure-virtual does ***NOT***
// mean the function can have no body!
class YouDontSeeThisOften
{
public:
	virtual PureButHasBody() = 0
	{
		cout << "You don't see this often!" << endl;
	}
};
 
int main(int argc, char* argv[])
{
	// YouDontSeeThisOften a;
		// If you uncomment the line above you'll get a compiler error,
		// without it things are just fine.
 
	return 0;
}

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;
}