quick question on what's being printed

[jimmah4013]

Hopefully you can help me. So I know the answer is 1 and 2, but I am not fully understanding the main() section. What if I changed A* to B*, would I go to class B and start there first and print out 2 and then 1? Can someone please help me understand this.

Thank you

class A
{ public:
int i;
A ( int x ) { i=x; }
virtual void f() const { cout<<i<<endl; }
};
class B : public A
{ public:
B ( int x ) : A(x) {}
virtual void f() const { cout<<i*2<<endl; }
};
void print ( A a ) { a.f(); }
void print ( A* a ) { a‐>f(); }
int main()
{
A* a = new B(1);
print ( *a );
print ( a );
}

[SamuelAdams]

http://www.cplusplus.com/articles/z13hAqkS/

[doug4]

In your program, B is derived from A, which means that all Bs are As, but not all As are Bs (All teacher are people, but not all people are teachers.) When B derives from A, the class B has everything contained in A as well as specialization information. In this case, the specialization information is an override of the f() function.

Since a B object is also an A object, you can point to it with an A* thus:A* a = new B(1) is legitimate.

Anywhere you need to pass an A object, you can use a B object instead. What happens is that the base class bits of the B object are used, and the specialization information is ignored.

This leads us to the print ( *a ); line. What happens is that a object of class A is to be passed into the function. Because an object of type B is used, a temporary A object is created by copy constructor (from the sliced B object) and placed on the stack. This is an actual A object, not a B object as its originator, so the temporary object has A's version of the virtual f() function.

The print ( a ); statement passes a pointer to the B object to its function. The object does not have to change. The original object, constructed as a B, has B's version of the virtual f() function.

A tutorial page from this web site which might help you better understand:
http://cplusplus.com/doc/tutorial/polymorphism/

I hope this helps.

p.s. Code tags, as SamuelAdams pointed to, would help in the future.