At 12:06 AM +0000 12/10/01, rz0 wrote:

--- In Boost-Users@y..., Jon Kalb wrote:

...

At 2:56 AM +0000 12/9/01, rz0 wrote:

...

#include <iostream> #include

class A { public: virtual void func1() { std::cout << "A::func1" << std::endl; } };

class B : public A { public: void func1() { std::cout << "B::func1" << std::endl; } void funcB() { std::cout << "B::funcB" << std::endl; } };

class C : public A { public: void func1() { std::cout << "C::func1" << std::endl; } void funcC() { std::cout << "C::funcB" << std::endl; } };

class Holder { boost::shared_ptr<A> aptr; public: void setclass(boost::shared_ptr<A> a) { aptr = a; } void callfunc() { aptr->func1(); } };

int main() {

Holder holder;

boost::shared_ptr<B> b(new B()); boost::shared_ptr<C> c(new C());

holder.setclass(b); //<------- holder.setclass(c); //<-------

holder.callfunc();

return 0; }

VC++ 6.0 flags the above lines with a conversion error.

Is there any way to resolve this?

Can you just define b and c as boost::shared_ptr<A>?

Hi Jon

Thank you very much for the response. Unfortunately this will cause the same problem that I ran into with std::auto_ptr and that is methods funcB and funcC which are exclusive to B and C are not allowed to be called.

Your original code worked for me, but I'm using CodeWarrior. I noticed that the function that you want to have called: template<typename Y> shared_ptr(const shared_ptr<Y>&); is #ifdef'ed on BOOST_MSVC6_MEMBER_TEMPLATES, but I don't know what that define means. -- Jon Kalb Kalb@LibertySoft.com