Re: Constness with standard containters of pointers

From:
Kai-Uwe Bux <jkherciueh@gmx.net>
Newsgroups:
comp.lang.c++
Date:
Sat, 17 May 2008 05:38:04 -0400
Message-ID:
<482ea77c$0$22078$6e1ede2f@read.cnntp.org>
Javier wrote:

Hello,

I have some cuestions about constness with standard containers and
pointers.
Supose I have a list of pointers to some class B:

     std::list< B * > > list;

I have readed that constness in std::list is the same that it is in C
arrays (const std::list makes const both the list and its content):

     const std::list< B * > > constList;

I can't do:

     constList->push_back(x);

But, is it still possible to call any non-const member funcion of B?:

     void B::nonConstMemberFuncion();

As in:

     constList.begin()->nonConstMemberFuncion();

a) It is possible. I don't want the user to add elements nor modify
any of the list. Which is the correct way of doing this?

     const std::list< const B * const > > constList;
     const std::list< const B * > constList;

b) It is not possible. Then, I suppose that the next sentences are the
same:

     const std::list < B > constList;
     const std::list < const B > constList;

   If I have the next class:

     class A {
     public:
          std::list< B * > & list() { return m_list; };
     private:
          std::list< B * > m_list;
     };

   How I add a const version for A::list()?


Your problem is that the list contains B* object and making those const
(e.g. by refering to the list via a const reference) does not make the
pointees const. You could fix that using a smart pointer that forwards
constness:

template < typename T >
class const_forwarding_ptr {

  T * the_ptr;

public:

  const_forwarding_ptr ( T * ptr )
    : the_ptr ( ptr )
  {}

  T & operator* ( void ) {
    return ( *the_ptr );
  }

  T const & operator* ( void ) const {
    return ( *the_ptr );
  }

  T * operator-> ( void ) {
    return ( the_ptr );
  }

  T const * operator-> ( void ) const {
    return ( the_ptr );
  }

  operator T * ( void ) {
    return ( the_ptr );
  }

  operator T const * ( void ) const {
    return ( the_ptr );
  }

};

#include <list>

int main ( void ) {
  std::list< const_forwarding_ptr< int > > the_list;
  int * i_ptr = new int ( 5 );
  the_list.push_back( i_ptr );
  *(*(the_list.begin())) = 6;
  std::list< const_forwarding_ptr< int > > const & c_ref = the_list;
  // the following line does not compile:
  // *(*(c_ref.begin())) = 5;
}

What about if I have a smart_pointer instead of a C pointer?

     const std::list< boost::shared_ptr< B > > constList =
something();
     constList.begin()->nonConstMemberFunction();


Depends on the smart pointer. With regard to shared_ptr from C++0X, it does
not forward constness to the pointee but acts similar to a raw pointer.

Best

Kai-Uwe Bux

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