How to enable_shared_from_this of both parent and derived

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());
    }

This can be made more convenient by placing it in a base class (for reuse).

#include <memory>

template <class Base>
class enable_shared_from_base
  : public std::enable_shared_from_this<Base>
{
protected:
    template <class Derived>
    std::shared_ptr<Derived> shared_from_base()
    {
        return std::static_pointer_cast<Derived>(shared_from_this());
    }
};

and then deriving from it as follows.

#include <functional>
#include <iostream>

class foo : public enable_shared_from_base<foo> {
    void foo_do_it()
    {
        std::cout << "foo::do_it\n";
    }
public:
    virtual std::function<void()> get_callback()
    {
        return std::bind(&foo::foo_do_it, shared_from_base<foo>());
    }
};

class bar1 : public foo {
    void bar1_do_it()
    {
        std::cout << "bar1::do_it\n";
    }
public:
    virtual std::function<void()> get_callback() override
    {
        return std::bind(&bar1::bar1_do_it, shared_from_base<bar1>());
    }
};