Re: Conditionally initializing a const reference without making a copy

From:

"Alf P. Steinbach" <alfps@start.no>

Newsgroups:

comp.lang.c++

Date:

Thu, 12 Apr 2007 00:26:42 +0200

Message-ID:

<5855l5F2euvstU1@mid.individual.net>

* JurgenvonOerthel@hotmail.com:

Consider the classes Base, Derived1 and Derived2. Both Derived1 and
Derived2 derive publicly from Base.

Given a 'const Base &input' I want to initialize a 'const Derived1
&output'.
If the dynamic type of 'input' is Derived1, then 'output' should
become a reference to 'input'.
Otherwise 'output' should become a reference to the (temporary) result
of the member function 'input.to_der1()' which returns a Derived1
object by value.

You can't conditionally hold on to temporary.

But you can do

   void foo( Derived1 const& o ) { ... }

   void bar()
   {
       ...
     if( dynamic_cast<Derived1 const*>( &input ) )
     {
         foo( input );
     }
     else
     {
         foo( dynamic_cast<Derived2 const&>( input ).to_der1() );
     }

It would probably be much better to make foo a virtual member function
of the base class.

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