问题描述

我有简单的基类和派生类，我希望它们都具有 shared_from_this().

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

这个简单的解决方案:

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());
    }
};

在以下代码中导致异常tr1::bad_weak_ptr:

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();

所以在谷歌搜索"之后，我找到了以下解决方案:

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());
    }
};

现在它可以工作了.

有没有更好、更方便、更正确的方法来为父母和孩子enable_shared_from_this?

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

谢谢

推荐答案

抱歉，没有.

问题在于 shared_ptr 和 shared_ptr 是不同的类型.我不了解幕后发生的一切，但我认为当构造函数返回并分配给 shared_ptr 时，内部 weak_ptr 看到没有任何东西指向它(因为只有 shared_ptr 会增加计数器)并重置自己.当您在 get_callback 中调用 bar1::shared_from_this 时，您会收到异常，因为内部 weak_ptr 没有指向任何内容.

The problem is that shared_ptr<foo> and shared_ptr<bar1> are different types. I don't understand everything that's going on under the hood, but I think that when the constructor returns and is assigned to a shared_ptr<foo>, the internal weak_ptr<bar1> sees that nothing is pointing to it (because only a shared_ptr<bar1> would increment the counter) and resets itself. When you call bar1::shared_from_this in get_callback, you get the exception because the internal weak_ptr isn't pointing to anything.

本质上，enable_shared_from_this 似乎只能从层次结构中的单个类透明地工作.如果您尝试手动实现，问题应该变得很明显.

Essentially, enable_shared_from_this only seems to work transparently from a single class in a hierarchy. If you try implementing it manually, the problem should become obvious.

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