How to make auto_ptr work polymorphically?

Check the following code,

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;
}

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

Code:

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auto_ptr<B> ptr(new D);

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

After calling the function, do the following:

Code:

---

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.

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:

Code:

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auto_ptr<B> ptr(new D);

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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.

Quote:

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Originally Posted by Philip Nicoletti

Code:

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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:

Code:

---

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.

Quote:

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Originally Posted by 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:

Code:

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auto_ptr<B> ptr(new D);

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and that would be fine.

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That is a good idea. Thanks.

Quote:

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Originally Posted by LarryChen

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.

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

Quote:

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Originally Posted by 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.

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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

Quote:

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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>.

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Quote:

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Originally Posted by D_Drmmr

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

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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

Code:

---

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);
}

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

Code:

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

        bar( auto_ptr<B>(ptr) );
}

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