Virtual Functions?

msr_codeproject

class Base
{
public:
Base() { }
virtual void foo1() { }
virtual void foo2() { }

} ;

class Derived : public Base
{
public:
Derived() { }
void foo1() { }
void foo3() { }
} ;

int _tmain(int argc, _TCHAR* argv[])
{
Derived * d = (Derived*)new Base() ;
d->foo3() ;
return 0;
}

In above code, 1) How it is possible to call foo3() of derived class. 2) If I call d->foo1(), Base class foo1() is getting called. I am not able to understand this. Please explain. Regards msr

CPallini

d is initialized with the vtable of Base, that's the reason why Base::foo1 is called. On the other hand it is declared as a pointer to a Derived object, hence the compiler (you fooled) calls the foo3 method. Try

class Derived : public Base
{
int data;
public:
Derived() { data = 100;}
void foo1() { }
void foo3() { cout << data << endl;}
} ;

Too see why fooling the compiler is not a good idea.

Veni, vidi, vici.

msr_codeproject

Hi, I have tried this. I am getting some junk value. Why is this happened? :( Regards msr

David Crow

msr_codeproject wrote:

How it is possible to call foo3() of derived class.

Just like you have it. What's the problem? :confused:

"One man's wage rise is another man's price increase." - Harold Wilson

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CPallini

Because there is no instance of Derived, the method reads garbage.

Veni, vidi, vici.

Chuck OToole

OK, this is real simple. You instantiate an instance of the "Base" class and then you fool the compiler into thinking it's a "Derived" class object. Then you called a function that only exists in a "real" instantiation of a "Derived" class object. What you did was fool yourself into expecting something reasonable to happen and you're surprised when it doesn't.

jschell

msr_codeproject wrote:

I am not able to understand this. Please explain.

In addition to the other replies... Don't write code like that. It is bad code.

JackDingler

Bad line here below. Don't do that.

Derived * d = (Derived*)new Base() ;

Try

Derived * d = new Derived;