Re: A Sample auto_ptr implementation
On Oct 13, 2:58 pm, Kai-Uwe Bux <jkherci...@gmx.net> wrote:
Yakov Gerlovin wrote:
Try to define assignment operator without template U
parameter:
//assignment operator
MyAutoPtr<T>& operator=(MyAutoPtr<T>& rhs);
Why do you need that 'U' anyway, release should return the
pointer to be passed to reset, so T and U should be equal.
He needs it to support polymorphism. When D is derived from B,
MyAutoPtr<B> should be constructible and assignable from
MyAutoPtr<D>.
But the OP could define copy-constructor and assignment
operator from T separately. For the copy-constructor, I recall
that this is needed. I am not sure if it is needed for the
assignment operator.
If I understand the context correctly, it is. A template
operator= will never stop the compiler from generating its
default copy operator=.
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