What happens in this program ?

[binarylife]

Please refer the following code :

#include<iostream>
using namespace std;
class abc
{
	private:int a;
	public: abc()
	{
		a=7;
	}
	public:void display()
	{
		cout<<a;
	}
};

class bcd: public abc
{

};


int main()
{
	
	bcd b1;
	b1.display();
	
	
}

It gives the output 7 .

looks like "b1.display()" is actually executing "abc::display()"
???

[Albatross]

@Question:
Yes, it does. Why wouldn't it? bcd inherits abc and doesn't overload display().

Lines 5, 6, and 10:
Naw, you don't need to have the access specifiers in the same line as each declaration, like in Java. This is perfectly valid:

class abc
{
private:
	int a;
public:
	abc() {
		a=7;
	}
	void display() {
		cout<<a;
	}
};

EDIT: -Albatross

[binarylife]

You mean, if in case a derived function is not overridden, base class function will be called always?

[MikeyBoy]

Yes. That's what inheritance is!

[binarylife]

thnx !