Virtual functions and polymorphism

Question

Suppose I have this:

class A
{
    public:
    virtual int hello(A a);
};

class B : public A
{
   public:
   int hello(B b){ bla bla };
};

So,

Answer 1

What you are doing there is overloading not overriding, i.e. it's as if class B is:

class B
{
public:
  int hello(A a) {...}
  int hello(B b) {...}
};

You have two functions of the same name, but with different signatures, which makes them different functions (just like the standard library has different functions for abs(float) and abs(double) etc.)

If you want to override, then you need to have the same signature, i.e. class B's hello needs to take a parameter of type A. That way, when you call hello on an object of class B, it will use class B's hello rather than class A's.

If you actually want class B's hello to only accept objects of type B then what you have is fine, although you probably want to make class A's hello non-virtual as you are not really wanting to override it -- you are defining a new function with new parameters and new behaviour.

Answer 2

Thansk for the answers, but I have to clarify some things to get my final answer.

Suppose I have the A class exactly how I defined it in the original question. And I add another method:

class A {
    ...
    int yeah();
}

Then I define class B as the following:

class B : public A {
    int hello(A a);
};

And another class C analogous to B.

What I know because I'm the programmer, it's that the hello methods of B and C are gonna have obviously A type objects as parameters, but instances of the same class. In example:

B b; 
b.hello(some_other_b_instance);

or

C c; 
c.hello(some_other_c_instance);

The problem is that in each hello function of the classes B and C, I want to do particular things with atributes of the particular class B or C. And because of the parameter is of type A, I cannot use them.

What I would need it's a kind of inverse polymorphysm, but its wrong because by definition I can send a C instance to B hello class, but I know it's not gonna happend.

I hope you get the idea of the code... A clase is abstract, and the real work makes sense in the particular clases B and C, each one do the work in their particular way to make the yeah function work. But B and C need to access their members to do the hello work correctly.

Answer 3

A bunch of good answers telling you WHAT happens, I thought I'd jump in with WHY.

There's this thing called the Liskov Substitution Principle, which says that the function in the subclass has to work under the same preconditions and postconditions as the base class. In this case, the function has to be able to operate on any object of type A. Note that because of the inheritance relationships, every B is-a A, but not every A is-a B. So to replace the base method, the new function in the derived class can weaken the preconditions or strengthed the postconditions, but not strengthen preconditions or weaken postconditions.

Your attempt to override strengthens the precondition, it accepts Bs, not all As.

Note that covariance IS allowed on return types. If your base class returned A, then it guarantees that the return value is-a A. The base class could then return a B, because every B is-a A.

But for input parameters, only contravariance meets the theoretical requirements of the LSP, and in/out parameters are invariants. In C++ in particular, all parameter types are invariant for the purposes of overloading.

Answer 4

When you override a method, it redefines what the method will do. You can only override virtual members that are already defined (with their set of parameters). If the type is of A, the method on A will be called. If the type is of B, the method on B will be called even if the variable is typed A but contains an instance of type B.

You can't change the parameter definitions for an overridden method, or else it would cease to be an override.

Answer 5

Your class B doesn't override the member function in A, it overloads it. Or tries to anyway, see the bit about hiding later.

Overriding is when a derived class defines its own version of a virtual member function from a base class. Overloading is when you define different functions with the same name.

When a virtual call is made on a pointer or reference that has the type of the base class, it will only "consider" overrides in the derived class, not overloads. This is essential - for an instance of B to be treated by callers as though it does everything an A can do (which is the point of dynamic polymorphism and virtual functions), its hello function needs to be able to take any object of type A. A hello function which only takes objects of type B, rather than any A, is more restrictive. It can't play the role of A's hello function, so it's not an override.

If you experiment a bit with calling hello on A and B, passing objects of type A or B, you should be able to see the difference. A has a function taking an A (which you haven't defined, so if you call it then your program will fail to link, but you can fix that). B has a function taking a B. They happen to have the same name, and of course since B derives from A, you can pass a B to the function taking an A. But B's function doesn't act as an override in virtual calls.

It is possible to call A's function on a B object, but only via a reference or pointer to A. A feature of C++ is that the definition of hello in B hides the definition in A. If overloading is what you want, it's possible to un-hide the base class function by adding using A::hello; to class B. If overriding is what you want, you have to define a function taking the same parameters. For example:

#include <iostream>

class A
{
    public:
    virtual int hello(A a) {std::cout << "A\n"; }
    virtual int foo(int i) { std::cout << "A::Foo " << i << "\n"; }
};

class B : public A
{
   public:
   using A::hello;
   // here's an overload
   int hello(B b){ std::cout << "B\n"; };
   // here's an override:
   virtual int foo(int i) { std::cout << "B::Foo " << i << "\n"; }
};

int main() {
    A a;
    B b;
    a.hello(a);  // calls the function exactly as defined in A
    a.hello(b);  // B "is an" A, so this is allowed and slices the parameter
    b.hello(a);  // OK, but only because of `using`
    b.hello(b);  // calls the function exactly as defined in B
    A &ab = b;   // a reference to a B object, but as an A
    ab.hello(a); // calls the function in A
    ab.hello(b); // *also* calls the function in A, proving B has not overridden it
    a.foo(1);    // calls the function in A
    b.foo(2);    // calls the function in B
    ab.foo(3);   // calls the function in B, because it is overridden
}

Output:

A
A
A
B
A
A
A::Foo 1
B::Foo 2
B::Foo 3

If you take away the using A::hello; line from B, then the call b.hello(a); fails to compile:

error: no matching function for call to `B::hello(A&)'
note: candidates are: int B::hello(B)

Answer 6

First, A is not an abstract class in your code. It must have at least one pure virtual function to be abstract.

1. different parameters means completely different method, even though the name is the same. Think of it as a different name. That's why it's called "signature". If A would be an abstract class, this code would not compile at all.

2. A::hello() will be called. No problem with that, and parameter must be type A, as if there was no inheritance.