How to make auto_ptr work polymorphically?

[LarryChen]

Check the following code,

class B
{
public:
	virtual void display()
	{
		cout<<"B"<<endl;
	}
};

class D : public B
{
public:
	void display()
	{
		cout<<"D"<<endl;
	}
};

void bar(auto_ptr<B> ptr)
{
	ptr->display();
}

int main(int argc, char* argv[])
{
    auto_ptr<D> ptr(new D);

	bar(ptr);
	return 0;
}

Actually it won't compile. I wonder how to make it work? Thanks.

[Philip Nicoletti]

auto_ptr<B> ptr(new D);

Note : passing atuo_ptrs are problematic because of the ownership
of the pointer when copied.

After calling the function, do the following:

bar(ptr);

if (ptr.get() == NULL)  
{
   cout << "null\n"; 
}

Also, add a destructor to the classes, printing out a message, and see
when the pointer gets deleted.

[Lindley]

As with all templates, a class<A> is not related to a class<B>, no matter whether A and B are related or not.

However, you could write:

auto_ptr<B> ptr(new D);

and that would be fine.

Note, however, that the D object will be destroyed after the call to bar() since you're passing the auto_ptr by value. This is unintuitive, which is why auto_ptr has been deprecated in favor of unique_ptr, which has the same behavior but is explicitly only movable, not copyable.

[LarryChen]

auto_ptr<B> ptr(new D);

Note : passing atuo_ptrs are problematic because of the ownership
of the pointer when copied.

After calling the function, do the following:

bar(ptr);

if (ptr.get() == NULL)  
{
   cout << "null\n"; 
}

Also, add a destructor to the classes, printing out a message, and see
when the pointer gets deleted.

I tried to call ptr.get() but it doesn't return NULL after passing to the function bar. I expect it to be NULL but I don't know why it isn't. Thanks.

[LarryChen]

As with all templates, a class<A> is not related to a class<B>, no matter whether A and B are related or not.

However, you could write:

auto_ptr<B> ptr(new D);

and that would be fine.

That is a good idea. Thanks.

[Philip Nicoletti]

I tried to call ptr.get() but it doesn't return NULL after passing to the function bar. I expect it to be NULL but I don't know why it isn't. Thanks.

What compiler/version are you using ? auto_ptr in VC++ version 6
had bugs in the implementation that was shipped.

[D_Drmmr]

As with all templates, a class<A> is not related to a class<B>, no matter whether A and B are related or not.

But if you write your own template class you can make class<A> convertible to class<B> by providing an overloaded constructor, as is done for shared_ptr.

From http://www.boost.org/doc/libs/1_44_0...shared_ptr.htm

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>.

[superbonzo]

But if you write your own template class you can make class<A> convertible to class<B> by providing an overloaded constructor, as is done for shared_ptr.

From http://www.boost.org/doc/libs/1_44_0...shared_ptr.htm

in principle auto_ptr can convert from/to auto_ptr's whose base pointers are convertible;

the problem in the OP code is that auto_ptr exposes both a template copy ctor taking any auto_ptr<T> by non const reference and a template conversion operator to any auto_ptr<T> which give an ambiguous overload resolution error in the bar argument copy-initialization. As in

struct A;

struct B { B( A& ); }; // if it were "B( const A& );" it would compile

struct A { operator B(); };

void bar( B );

int main()
{
	A a;
	bar(a);
}

indeed, an alternative to "auto_ptr<B> ptr = auto_ptr<B>( new D() )" is

int main()
{
	auto_ptr<D> ptr(new D);

	bar( auto_ptr<B>(ptr) );
}