Question

Declarations of auto_ptr from C++ Standard Library

namespace std {

template <class Y> struct auto_ptr_ref {};


template <class X>
class auto_ptr {
public:
    typedef X element_type;

    // 20.4.5.1 construct/copy/destroy:
    explicit           auto_ptr(X* p =0) throw();
                       auto_ptr(auto_ptr&) throw();
    template <class Y> auto_ptr(auto_ptr<Y>&) throw();

    auto_ptr&                      operator=(auto_ptr&) throw();
    template <class Y> auto_ptr&   operator=(auto_ptr<Y>&) throw();
    auto_ptr&                      operator=(auto_ptr_ref<X>) throw();

    ~auto_ptr() throw();

    // 20.4.5.2 members:
    X&     operator*() const throw();
    X*     operator->() const throw();
    X*     get() const throw();
    X*     release() throw();
    void   reset(X* p =0) throw();

    // 20.4.5.3 conversions:
                                auto_ptr(auto_ptr_ref<X>) throw();
    template <class Y> operator auto_ptr_ref<Y>() throw();
    template <class Y> operator auto_ptr<Y>() throw();
};

}

I don’t understand the purpose of this part:

template <class Y> struct auto_ptr_ref {};

Without declaring any variable, how can these be valid:

auto_ptr&                      operator=(auto_ptr_ref<X>) throw();

and these too:

auto_ptr(auto_ptr_ref<X>) throw();
    template <class Y> operator auto_ptr_ref<Y>() throw();

Edit: and also (I just notice) I don’t understand how the “operator” are used for the last two lines. Isn’t the syntax something like “return-type operatoroperand;”, where is the return type? operand?

Answer 1

Google search for “auto_ptr_ref” reveals this detailed explanation.

I don’t quite get that explanation, but looks like it is for the following. Without that trick you could pass an auto_ptr into a function that would get ownership of the object and assign your variable to a null pointer. With the extra class trick above you will get a compile error in such case.