Re: How to use static_cast for std::list of derived objects?

From:
benben <benhonghatgmaildotcom@nospam.invalid>
Newsgroups:
comp.lang.c++.moderated
Date:
21 Jan 2007 10:54:53 -0500
Message-ID:
<45b2d456$0$9771$afc38c87@news.optusnet.com.au>
Javier Lopez wrote:

 I would like to write a function with an std::list of a base class as
 argument. I want to call this function making a cast of derived
classes
 from this base class. How can use the static_cast to do that?

 Here I show you some code that illustrates what I want to do.


[trivial code snipped]

 void func(list<A>& a_list)
 {
   for (list<A>::iterator iter = a_list.begin() ; iter != a_list.end()
; iter++)
     cout << iter->m_value << ' ';
   cout << endl;

 }

 int main()
 {

   B b;
   list<B> b_list;
   for (int i=0 ; i<10 ; i++)
     {
       b.m_value = i;
       b_list.push_back(b);
     }

   func((list<A>&)(b_list)); // Compile and works
   func(reinterpret_cast<list<A>&>(b_list)); // Also compile and works
   func(static_cast<list<A>&>(b_list)); // Does not compile (see below)
 }


How about:

    void func(const list<A*>& a_list);

    int main()
    {
        list<B> b_list;
        list<A*> a_list;

        B b;

        for (int i = 0; i < 10; ++i)
        {
            b.m_value = i;
            b_list.push_back(b);
            a_list.push_back(&(b_list.back()));
        }

        func(a_list);

    }

Or better, following the STL spirit:

    template <typename Itr>
        void func(Itr begin, Itr end)
        {
            for (Itr i = begin; i != end; ++i)
                cout << i->m_value << " ";

            cout << endl;
        }

    int main()
    {
        list<B> b_list;

        for ... // fill up b_list just like you did

        func(b_list.begin(), b_list.end());
    }

 The last function call does not compile with gcc 4.1.1 and give me the
 following output:

 error: invalid static_cast from type 'std::list<B, std::allocator<B>

' to type 'std::list<A, std::allocator<A> >&'


 Could someone explain me why?


Well, list<B> is not really a list<A>, neither is list<B*> a list<A*>.
Why? Because...

           http://www.research.att.com/~bs/bs_faq2.html#conversion

 Thank you very much in advance.


Ben

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