Re: how to reuse a copy constructor in an operator = function

From:

"Andrew Koenig" <ark@acm.org>

Newsgroups:

comp.lang.c++

Date:

Sun, 04 Nov 2007 16:34:56 GMT

Message-ID:

<QwmXi.51066$kj1.37149@bgtnsc04-news.ops.worldnet.att.net>

"Kai-Uwe Bux" <jkherciueh@gmx.net> wrote in message
news:fgii3a$mpu$1@murdoch.acc.Virginia.EDU...

It also causes no end of problems if someone inherits from the
class.

(a) That's one of the gotchas, and

(b) the reason for using

this->T::~T();

instead of

this->~T();

That actually takes care of _many_ problems this method can cause in the
context of inheritance.

Well yes, but it misses the most important problem.

Suppose you derive from T, and write an assignment operator in the derived
class that calls the one from the base class:

    class T1: public T {

    // ...
        T1& operator=(coust T1& x) {
            this->T::operator=(x);
            // Do some other stuff here
            return *this;
        }

    // ...
    };

When T1::operator= calls T::operator=, T::operator= will stealthily destroy
the object, which had dynamic type T1, and replace it with a new object that
has dynamic type T. Of course, the compiler won't know about this for the
purpose of static type checking.

The result will be chaos.