Re: problems with polymorphism and inheritance with templates

From:
SG <s.gesemann@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Mon, 30 Aug 2010 17:59:49 CST
Message-ID:
<9951fcaa-75d2-4633-aa69-8893fdb677d2@s9g2000yqd.googlegroups.com>
On 29 Aug., 17:47, Frank Buss wrote:

This code:

  template<class T>
  class Foo
  {
  public:
        T* member;
  };


Ok, let's assume Foo<T> has a default constructor that initializes
'member' with a null pointer.

  class Base
  {
  };

  typedef Foo<Base> BaseFoo;

  class Derived
  {
  };


It seems, you forgot to add Base as a base class for Derived here.

  typedef Foo<Derived> DerivedFoo;

  int main(int argc, char** argv)
  {
     DerivedFoo* dfoo = new DerivedFoo();
     BaseFoo* bfoo = dfoo;
     return 0;
  }


Why do you 'new' DerivedFoo here?

doesn't compile, the compiler says "cannot convert ?DerivedFoo*? to
?BaseFoo*? in initialization".


DerivedFoo and BaseFoo are two distinct types that have no inheritence
relationship. But you can make a DerivedFoo convertible to a BaseFoo
and this is probably what you really wanted to do:

   int main(int argc, char** argv)
   {
      DerivedFoo dfoo;
      BaseFoo bfoo = dfoo;
      return 0;
   }

You can do this by providing a templated conversion constructor in the
Foo class template:

   template<class T>
   class Foo
   {
   public:
      T* member;
      Foo() : member(0) {} // default ctor

      template<class U>
      Foo(Foo<U> const& f) : member(f.member) {}
   };

As a bonus you might want to disable this templated conversion
constructor in case U* is not convertible to T* for overloading
reasons. It's possible with a bit of meta-programming but I think this
would be too much information for now.

I know that something similiar (but not really the same) is possible,
because shared_ptr can do it, e.g. this one compiles with g++ 4.3.2:
  [...]
  typedef shared_ptr<Base> BasePtr;
  typedef shared_ptr<Derived> DerivedPtr;

int main(int argc, char** argv)
{
    DerivedPtr dptr(new Derived());
    BasePtr bptr = dptr;
    return 0;
}


Right. The thing is DerivedPtr is convertible to BasePtr but the types
are not reference-related because there is no inheritence relationship
between these two types. And since they are not reference-related you
cannot use a pointer of type BasePtr* to point to an object of type
DerivedPtr.

Maybe the typedefs are confusing you. You are aware of the fact that
BasePtr* is a pointer to a struct containing a pointer to Base, right?
The inheritence relationship is only between Base and Derived, not
between BasePtr and DerivedPtr. Is it possible that you missed one
level of indirection?

Cheers!
SG

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