Re: problems with polymorphism and inheritance with templates

=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <>
Mon, 30 Aug 2010 08:35:34 CST
On 29 Aug., 17:47, Frank Buss <> wrote:

This code:

template<class T>
class Foo
       T* member;

class Base

typedef Foo<Base> BaseFoo;

class Derived

In consideration of what you wrote later combined with your
chosen naming conventions, I assume this was actually
meant to be:

class Derived : public Base

typedef Foo<Derived> DerivedFoo;

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

doesn't compile, the compiler says "cannot convert ?DerivedFoo*? to
?BaseFoo*? in initialization". How can I fix it? Of course, this one works:

       Derived* derived = new Derived();
       Base* base = derived;

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:

#include <tr1/memory>

using namespace std::tr1;

class Base

typedef shared_ptr<Base> BasePtr;

class Derived : public Base

typedef shared_ptr<Derived> DerivedPtr;

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

It is quite easy to simulate a behaviour that follows
the strategy of shared_ptr, unique_ptr, and similar
classes that can reasonably accept specializations
for different types than the own one. But it is impossible
to "simulate" a base-derived relationship without
introducing such a relation.

The first question is: Why do you need the template,
if you want a specific class hierarchy? Either Foo is
supposed to be some form of a smart pointer, then
there is no reason to create objects in the free store
and to require base-derived relations among smart
pointers. Or Foo is something else, but the question
is: What does it model?

HTH & Greetings from Bremen,

Daniel Kr?gler

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