Re: virtual assignment operator/polymorphism question

"James Kanze" <>
13 Apr 2007 02:18:27 -0700
On Apr 12, 1:15 pm, wrote:

I have a question following up the following slightly older


class Base
     virtual Base& operator = (const Base &k) {}

class Derived: public Base
     virtual Base& operator = (const Base &k) {}

int main(int argc, char* argv[])
     Derived *dp1 = new Derived();
     Derived *dp2 = new Derived();

     *dp1 = *dp2; // Base::operator= is called

How do you know? I think rather that the compiler generated
derived operator is being called (statically, since it isn't
virtual). This operator, of course, calls the base operator=.

     Base *bp = *dp1;
     *bp = *dp2; // Derived::operator= is called

Yes, but not the default Derived::operator=.

     return 0;

While it seems clear to me why *bp = *dp2 leads to the
Derived::operator= being called I do not understand why *dp1 = *dp2
calls the Base::operator=.
What's going on here???

Your Derived::operator= is not a copy assignment operator (since
it doesn't assign a Derived to Derived, but a Base to Derived),
so it doesn't inhibit the compiler from generating its version.

More generally, polymorphism and assignment don't work well
together. If you plan on using a class as a base, it's
generally best to declare the operator= private, and not
implement it (or to derive it from something like

James Kanze (GABI Software)
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