c++ - How to enable_shared_from_this of both parent and derived

Question

I have simple base and derived class that I want both have shared_from_this().

This simple solution:

class foo : public enable_shared_from_this<foo> {
    void foo_do_it()
    {
        cout<<"foo::do_it
";
    }
public:
    virtual function<void()> get_callback()
    {
        return boost::bind(&foo::foo_do_it,shared_from_this());
    }
    virtual ~foo() {};
};

class bar1 : public foo , public enable_shared_from_this<bar1> {
    using enable_shared_from_this<bar1>::shared_from_this;
    void bar1_do_it()
    {
        cout<<"foo::do_it
";
    }
public:
    virtual function<void()> get_callback()
    {
        return boost::bind(&bar1::bar1_do_it,shared_from_this());
    }
};

Causes exception tr1::bad_weak_ptr in following code:

shared_ptr<foo> ptr(shared_ptr<foo>(new bar1));
function<void()> f=ptr->get_callback();
f();

So after "googling" I have found following solution:

class bar2 : public foo {
    void bar2_do_it()
    {
        cout<<"foo::do_it
";
    }
    shared_ptr<bar2> shared_from_this()
    {
        return boost::static_pointer_cast<bar2>(foo::shared_from_this());
    }
public:
    virtual function<void()> get_callback()
    {
        return boost::bind(&bar2::bar2_do_it,shared_from_this());
    }
};

And now it works.

Is there any better and more convinient and correct way to enable_shared_from_this for both parent and child?

Thanks

Answer 1

The OP solution can be made more convenient by defining the following on the base class.

protected:
    template <typename Derived>
    std::shared_ptr<Derived> shared_from_base()
    {
        return std::static_pointer_cast<Derived>(shared_from_this());
    }