Virtual operator overloads don't seem to work?

S

Stuart Brockman

Hi,
I don't quite get what is going on in this code example:

---------------------------
#include <iostream>

using namespace std;

class Base{
public:
virtual void Test(){
cout << "Base" << endl;
}

virtual bool operator==(const Base &other){
cout << "Base Comparison" << endl;
return false;
}
};

class Derived : public Base{
public:
void Test(){
cout << "Derived" << endl;
}

bool operator==(const Derived &other){
cout << "Derived Comparison" << endl;
return true;
}
};

int main(int argc, char** argv) {
Base a; //Create a base object
a.Test(); //Outputs "Base" as expected
Derived b, c; //Create two derived objects
b.Test(); //Outputs "Derived" as expected
if(b==c) cout << "True" << endl; //Does derived comparison and
returns true as expected.
Base *d=&b, *e=&c; //Create two base pointers to derived objects
d->Test(); //Outputs "Derived" as expected
if(*d==*e) cout << "True" << endl; //Does base comparison and
returns false!?
return 0;
}
----------------------------
The output is:

Base
Derived
Derived Comparison
True
Derived
Base Comparison

Notice, that the line "d->Test()" works correctly, but the comparison
on the next line does not. The compiler (g++ (GCC) 4.2.3 (Ubuntu
4.2.3-2ubuntu7) ) seems to be ignoring the virtual-ness of
Base::eek:perator== . Is this correct? Have I made a mistake?
 
S

Stuart Brockman

That's because "bool operator==(const Derived &)" does not polymorphically
overload "bool operator==(const Base &)". In a derived class, the function's
signature must match the base class's virtual function, in order for it to
be polymorphically overloaded (the function parameters must match). In your
case, above, you have two different functions, no different that void foo()
and void bar(). One does not overload the other.


Yes, but a very natural one.

 application_pgp-signature_part
< 1KViewDownload

Ahh... I see... So what should I change Derived::eek:perator== to?
Clearly it would have to have a signature of bool
Derived::eek:perator==(const Base &other), but how do I make sure that I
get correct behavior when doing a "Derived==Base" (I would prefer that
Base::eek:perator== was called in this case)?
 
J

James Kanze

I don't quite get what is going on in this code example:
using namespace std;
class Base{
public:
virtual void Test(){
cout << "Base" << endl;
}
virtual bool operator==(const Base &other){
cout << "Base Comparison" << endl;
return false;
}
};
class Derived : public Base{
public:
void Test(){
cout << "Derived" << endl;
}
bool operator==(const Derived &other){
cout << "Derived Comparison" << endl;
return true;
}
};
int main(int argc, char** argv) {
Base a; //Create a base object
a.Test(); //Outputs "Base" as expected
Derived b, c; //Create two derived objects
b.Test(); //Outputs "Derived" as expected
if(b==c) cout << "True" << endl; //Does derived comparison and
returns true as expected.
Base *d=&b, *e=&c; //Create two base pointers to derived objects
d->Test(); //Outputs "Derived" as expected
if(*d==*e) cout << "True" << endl; //Does base comparison and
returns false!?
return 0;}
Base
Derived
Derived Comparison
True
Derived
Base Comparison
Notice, that the line "d->Test()" works correctly, but the
comparison on the next line does not. The compiler (g++ (GCC)
4.2.3 (Ubuntu 4.2.3-2ubuntu7) ) seems to be ignoring the
virtual-ness of Base::eek:perator== . Is this correct? Have I
made a mistake?

How could it do otherwise? Base::eek:perator==() has a different
signature than Derived::eek:perator==(), so the function in the
derived class hides, rather than overloads, the function in the
base class. Moreover, there's no way you could possibly call
Derived::eek:perator==(), since it requires a Derived const&, and
can't be called with a Base const& (which is basically what you
get when you dereference a Base const*).

In general, binary operators don't work very well with
inheritance; what you'd really need to make them work is double
dispatch. In the special case of == and !=, you can do
something like:

bool
Derived::eek:perator==( Base const& other ) const
{
Derived const* p
= dynamic_cast< Derived const* >( &other ) ;
return p != NULL && operator==( *p ) ;
}

in addition to your normal Derived::eek:perator==, at least if you
decide that two different derived types can never compare equal.
If it makes sense for two different derived types to compare
equal, however (say because they both represent the same value,
but in different ways), you'll need to set up a much more
complicated mechanism.
 
B

Bo Persson

Stuart said:
Ahh... I see... So what should I change Derived::eek:perator== to?
Clearly it would have to have a signature of bool
Derived::eek:perator==(const Base &other), but how do I make sure that
I get correct behavior when doing a "Derived==Base" (I would prefer
that Base::eek:perator== was called in this case)?

You can do that with a free function:

bool operator==(const Derived& Left, const Base& Right)
{ return Right.operator==(Left); }


Bo Persson
 

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