Virtual operator overloads don't seem to work?

[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==(cons t Base &other){
cout << "Base Comparison" << endl;
return false;
}
};

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

bool operator==(cons t 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::operator= = . Is this correct? Have I made a mistake?

[James Kanze]

Re: Virtual operator overloads don't seem to work?

On Oct 23, 12:09 am, Stuart Brockman <stuartbrock... @gmail.comwrote :

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==(cons t Base &other){
cout << "Base Comparison" << endl;
return false;
}
};

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

bool operator==(cons t 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::operator= = . Is this correct? Have I
made a mistake?

How could it do otherwise? Base::operator= =() has a different
signature than Derived::operat or==(), 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::operat or==(), 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::operat or==( Base const& other ) const
{
Derived const* p
= dynamic_cast< Derived const* >( &other ) ;
return p != NULL && operator==( *p ) ;
}

in addition to your normal Derived::operat or==, 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.

--
James Kanze (GABI Software) email:james.kan ze@gmail.com
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