Why "const" break the Polymorphism?

[July]

Hello!
consider the following code:

class A {
public:
virtual void f() const{
cout << "A::f()" << endl;
}
};

class B : public A {
public:
void f() {
cout << "B::f()" << endl;
}
};

int main()
{
B b;
A *pa = &b;
pa->f();
}

the output is A::f()
neither b nor pa is const , why A::f() get called?

Will somebody explain this?

Thanks

 

[Luke Meyers]

class A {
public:
virtual void f() const;{
cout << "A::f()" << endl;
}
};

class B : public A {
public:
void f() {
cout << "B::f()" << endl;
}
};

int main()
{
B b;
A *pa = &b;
pa->f();
}

the output is A::f()
neither b nor pa is const , why A::f() get called?

B::f() does not override A::f(), because the const modifier makes the
signatures different. It's the same as if you had different argument
lists.

Luke

 

[Guest]

Will somebody explain this?

Compare those two signatures:

virtual void A::f() const;
void B::f();

or without virtual keyword for better understanding:
void A::f() const;
void B::f();

Do those signatures declare the same type of function?
Does B::f override A::f?

Cheers

 

[mlimber]

B::f() does not override A::f(), because the const modifier makes the
signatures different. It's the same as if you had different argument
lists.

IOW, B::f() is not polymorphically accessible to a pointer of type A
since B::f() does not override a virtual function of A. In fact, the
pointer pa can never call B::f() at all, no matter what the const
qualifiers are on pa.

On the other hand, it is curious that the compiler does not complain
about the cv-qualifiers of pa. I can't see why it would even let you
invoke A::f() since pa is non-const.

Cheers! --M

 

[Neelesh Bodas]

On the other hand, it is curious that the compiler does not complain
about the cv-qualifiers of pa. I can't see why it would even let you
invoke A::f() since pa is non-const.

I didnot get what you want to say. What is the problem in invoking a
const member function on a non-const object?

 

[Earl Purple]

Compare those two signatures:

virtual void A::f() const;
void B::f();

or without virtual keyword for better understanding:
void A::f() const;
void B::f();

Do those signatures declare the same type of function?
Does B::f override A::f?

No, in fact B::f hides A::f. That means if you have a const pointer or
reference to B and call f() it will fail to compile because it does not
inherit A::f() const
..

 

[mlimber]

I didnot get what you want to say. What is the problem in invoking a
const member function on a non-const object?

Sorry. I wasn't thinking clearly. Now retreating to screw head on
straight.

Cheers! --M

 

[Guest]

No, in fact B::f hides A::f. That means if you have a const pointer or
reference to B and call f() it will fail to compile because it does not
inherit A::f() const

That's what I was trying to explain.
Seems, I wasn't very clear.

Cheers

 

[Massimo Soricetti]

Earl Purple ha scritto:

No, in fact B::f hides A::f. That means if you have a const pointer or
reference to B and call f() it will fail to compile because it does not
inherit A::f() const

(noob question)
This means that a virtual function must never be declared const?
OR that if a virtual function is declared const, every function trying
to override it must be declared const also?

 

[Gavin Deane]

Earl Purple ha scritto:

(noob question)
This means that a virtual function must never be declared const?
OR that if a virtual function is declared const, every function trying
to override it must be declared const also?

The second one. There is absolutely no problem with const virtual
functions.

The issue is that the const modifier forms part of the function's type.
And for virtual function overriding to work, the type of the derived
class function must match the type of the base class function exactly.
This means the return types must be the same [*], the number and types
of the parameters must be the same and the const-ness and volatile-ness
must be the same.

[*] With the exception of covariant return types.
http://www.parashift.com/c++-faq-lite/virtual-functions.html#faq-20.8

Gavin Deane