6

I have a base class which derives from boost::enable_shared_from_this, and then another class which derives from both the base class and boost::enable_shared_from_this:

#include <boost/enable_shared_from_this.hpp>
#include <boost/shared_ptr.hpp>

using namespace boost;

class A : public enable_shared_from_this<A> { };

class B : public A , public enable_shared_from_this<B> {
public:
    using enable_shared_from_this<B>::shared_from_this;
};

int main() {
shared_ptr<B> b = shared_ptr<B>(new B());
shared_ptr<B> b_ = b->shared_from_this();

return 0;
}

This compiles but at runtime it gives

terminate called after throwing an instance of 'boost::exception_detail::clone_impl<boost::exception_detail::error_info_injector<boost::bad_weak_ptr> >'
  what():  tr1::bad_weak_ptr
Aborted

What is causing this, and is there some way around it?

EDIT:

What if I need something like this:

class A : public enable_shared_from_this<A> { };
class B : public enable_shared_from_this<B> { };    

class C : public A, public B, public enable_shared_from_this<C> {
public:
    using enable_shared_from_this<C>::shared_from_this;
};

such that A and B both need shared_from_this on their own (and one can't inherit it from the other), and C needs A, B, and shared_from_this?

4

2 回答 2

10

You shouldn't inherit from enable_shared_from_this more than once in a given inheritance chain.

In this case, you can leave the base class A inheriting from enable_shared_from_this, and have the derived class B return a shared_ptr<A> and then static_pointer_cast it to shared_ptr<B>.

Or as Omnifarious pointed out, you could have a function in B which does this for you. Although, rather than overloading shared_from_this() I would favour explicitly-named functions to minimise surprises for clients of the class:

#include <boost/enable_shared_from_this.hpp>
#include <boost/shared_ptr.hpp>

using boost::shared_ptr;

class A : public boost::enable_shared_from_this<A> { };

class B : public A {
public:
    using enable_shared_from_this<A>::shared_from_this;
    shared_ptr<B> shared_B_from_this() {
        return boost::static_pointer_cast<B>(shared_from_this());
    }
    shared_ptr<B const> shared_B_from_this() const {
        return boost::static_pointer_cast<B const>(shared_from_this());
    }
};

int main() {
    shared_ptr<B> b = shared_ptr<B>(new B);
    shared_ptr<B> b1 = boost::static_pointer_cast<B>(b->shared_from_this());
    shared_ptr<B> b2 = b->shared_B_from_this();
    return 0;
}
于 2012-10-09T00:10:02.043 回答
5

Here's how I would solve your problem:

#include <boost/enable_shared_from_this.hpp>
#include <boost/shared_ptr.hpp>

using namespace boost;

class virt_enable_shared_from_this :
   public enable_shared_from_this<virt_enable_shared_from_this>
{
 public:
   virtual ~virt_enable_shared_from_this() {}
};

template <class T>
class my_enable_shared_from_this : virtual public virt_enable_shared_from_this
{
 public:
   shared_ptr<T> shared_from_this() {
      return dynamic_pointer_cast<T>(virt_enable_shared_from_this::shared_from_this());
   }
};

class A : public my_enable_shared_from_this<A> { };

class B : public my_enable_shared_from_this<B> { };

class C : public A, public B, public my_enable_shared_from_this<C> {
 public:
   using my_enable_shared_from_this<C>::shared_from_this;
};

int main() {
   shared_ptr<C> c = shared_ptr<C>(new C());
   shared_ptr<C> c_ = c->shared_from_this();

   return 0;
}

This is painful and at least a bit ugly. But it works reasonably well, after a fashion. I think Fraser's idea of re-thinking your design is likely the better option.

于 2012-10-09T06:24:27.040 回答