4

Some smart pointer templates, such as boost::shared_ptr, may be instantiated with void to hold an arbitrary object:

http://www.boost.org/doc/libs/1_39_0/libs/smart_ptr/sp_techniques.html#pvoid

Below is a minimal scoped_ptr implementation. When instantiated with void, the compiler complains about an illegal "reference to void" being formed in the dereference operator. It seems the "substitution failure is not an error" (SFINAE) rule does not cover this situation.

How then is it possible to implement a scoped_ptr? In particular, is there an alternative to writing a template specialization? This would cause large code reduplication with a realistic smart pointer implementation.

#include <cstdlib>

template<typename T>
void destroy(T* ptr)
{
    delete ptr;
}

class scoped_ptr_impl_base
{
public:
    virtual ~scoped_ptr_impl_base() { }
};

template<typename T, typename F>
class scoped_ptr_impl : public scoped_ptr_impl_base
{
public:
    scoped_ptr_impl(T* ptr, F dtor)
        : m_ptr(ptr), m_dtor(dtor)
    {
    }

    virtual ~scoped_ptr_impl()
    {
        m_dtor(m_ptr);
    }

private:
    T* m_ptr;
    F m_dtor;
};

template<typename T>
class scoped_ptr
{
public:
    explicit scoped_ptr(T* ptr = 0)
        : m_ptr(ptr),
          m_impl(new scoped_ptr_impl<T, void (*)(T*)>(&destroy<T>))
    {
    }

    template<typename F>
    scoped_ptr(T* ptr, F dtor)
        : m_ptr(ptr),
          m_impl(new scoped_ptr_impl<T, F>(ptr, dtor))
    {
    }

    ~scoped_ptr()
    {
        delete m_impl;
    }

    T& operator*()
    {
        return *m_ptr;
    }

    T* operator->()
    {
        return m_ptr;
    }

private:
    T* m_ptr;
    scoped_ptr_impl_base* m_impl;

    scoped_ptr(const scoped_ptr&);
    scoped_ptr& operator=(const scoped_ptr&);
};

int main()
{
    scoped_ptr<void> p(std::malloc(1), std::free);
    // scoped_ptr.cpp: In instantiation of `scoped_ptr<void>':
    // scoped_ptr.cpp:76:   instantiated from here
    // scoped_ptr.cpp:56: error: forming reference to void
    // (g++ 4.3.3)

    return 0;
}
4

1 回答 1

8

You could use a type trait for the reference type:

template<typename T>
struct type_trait
{
    typedef T& reference;
};

template<>
struct type_trait<void>
{
    typedef void reference;
};

then in your scoped_ptr_impl :

typename type_trait<T>::reference operator*()
{
    return *m_ptr;
}

Not sure if void is the right type in the specialisation though . What type do you want it to return?

于 2009-08-21T15:24:00.103 回答