shared_ptr: 复制基数class的shared_ptr时引用计数会增加吗？

Question

shared_ptr<T> can be implicitly converted to shared_ptr<U> whenever T* can be implicitly converted to U*. In particular, shared_ptr<T> is implicitly convertible to shared_ptr<T const>, to shared_ptr<U> where U is an accessible base of T, and to shared_ptr<void>.

但我没有找到任何地方写的，如果这样做，它会增加引用计数器。

我尝试了以下代码，it works：

struct A {
    virtual int foo() {return 0;}
};

struct B : public A {
    int foo() {return 1;}
};

int main() {
    boost::shared_ptr<A> a;
    {
        boost::shared_ptr<B> b(new B());
        a = b;
    }

    std::cout << a->foo() << std::endl; ///Prints 1
}

假设情况总是如此，但我找不到可以证实这一点的信息来源。

Answer 1

好的，当我写完问题时，我终于找到了答案，在copy constructor documentation

shared_ptr(shared_ptr const & r); // never throws
template<class Y> shared_ptr(shared_ptr<Y> const & r); // never throws
Requires: Y* should be convertible to T*.

Effects: If r is empty, constructs an empty shared_ptr; otherwise, constructs a shared_ptr that shares ownership with r.

Postconditions: get() == r.get() && use_count() == r.use_count().

所以，答案是是。

shared_ptr: 复制基数class的shared_ptr时引用计数会增加吗？

shared_ptr: Does the reference count increase when copying in a shared_ptr of the base class?

c++

shared-ptr

c++98