Re: Does Virtual functionality works with assignment operator?
On Feb 9, 8:45 am, Kai-Uwe Bux <jkherci...@gmx.net> wrote:
Ian Collins wrote:
sukhpal wrote:
i have tried this following Example.
i am expecting the out-put as:
"in a my_first class assignment" but i am getting the out-put as
"in a base class assignment"
That's correct, you can't have a virtual assignment operator.
Well, it appears that you can, but it is useless:
Of course you can. It's not useless, but it's usually a design
error.
#include <iostream>
#include <ostream>
struct Base {
virtual
void operator= ( Base const & ) {
std::cout << "base\n";
}
};
struct Derived : public Base {
virtual
void operator= ( Base const & ) {
Note that this is an overload. It is NOT a copy assignment
operator, so the compiler will still generate its version of a
copy assignment operator (which will call the copy assignment
operator you defined in the base class).
std::cout << "derived\n";
}
};
void display ( Base & lhs, Base & rhs ) {
lhs = rhs;
}
int main ( void ) {
Base b;
Derived d;
b = b;
b = d;
d = b;
d = d;
display( b, b );
display( b, d );
display( d, b );
display( d, d );
}
Output:
base
base
derived
base
base
base
derived
derived
The most funny thing is line 4 where d=d prints "base", which
I do not understand.
The compiler has generated a copy assignment operator for
Derived. It's the best match here, so it's what gets called.
And it calls the base class assignment operator, which does the
output.
Note that in anything but a toy example like this, the results
probably wouldn't make sense. If I do something like: a = b, I
expect that a and b are "equivalent" in some sense after the
operation. In most cases, that equivalence should mean that
they have the same type. And there's no way you can implement
that. (In other words, there's a very real reason why "OO"
languages don't support value semantics.)
Is there undefined behavior? Standard anyone?
No undefined behavior. See =A712.8, and the conditions controling
when and how the compiler generates a copy assignment operator.
(Then overload resolution, but I think that once you realize
that Derived has two assignment operators, the rest will be
clear to you.)
--
James Kanze (GABI Software) email:james.kanze@gmail.com
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