Hello,

here are a few questions concerning auto_ptrs:

1) I read Scott Meyer's article on auto_ptr at
http://www.awl.com/cseng/titles/0-201-63371-X/auto_ptr.html,
and I don't agree with the discussion of the derived-to-base
direct-initialization case:

    struct Base {};
    struct Derived : Base {};
    auto_ptr<Derived> source();

    auto_ptr<Base> p( source() );

The GOTW #25 quoted in this article says that there is one viable
constructor:

    auto_ptr<Base>::auto_ptr(auto_ptr_ref<Base>),

callable using the conversion

    auto_ptr<Derived>::operator auto_ptr_ref<Base>().

However, it seems to me that the constructor actually is

    auto_ptr<Base>::auto_ptr(auto_ptr<Base>::auto_ptr_ref<Base>),

while the conversion operator is

    auto_ptr<Derived>::operator auto_ptr<Derived>::auto_ptr_ref<Base>().

auto_ptr<Base>::auto_ptr_ref<Base> and auto_ptr<Derived>::auto_ptr_ref<Base>
are different (and unrelated) types, right ?
If so, that means that there is no viable constructor, and that
direct-initialization from an auto_ptr of derived class is ill-formed.
For that trick to work, the auto_ptr_ref member template, should be
a member of a base class common to all auto_ptrs.


2) Consider:

struct A{};
auto_ptr<A> source();
auto_ptr<A> p;

p=source();

Since source() returns an rvalue, and operator='s parameter is a
reference to non-const, this is ill-formed. Was a operator=(auto_ptr_ref<X>)
forgotten ?


3) 20.4.5.3 does not require Y* to be implicitly convertible to X*
in the specification of

    template<class Y> operator auto_ptr<Y>().

But if an implementation makes the casts explicit, a compiler will
silently accept copy-initialization from base class to derived class.
This certainly is not the intent. Did I miss something ?

--
Emmanuel Lesueur - lesueur@desargues.univ-lyon1.fr


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