I've been looking through N2284 and was looking at queue on page 672
and found some strange inconsistencies.

In operator=(queue&&q),  std::move(q.c) is write. Later in
swap(queue&&q), the "std::move" is missing. Seeing as most containers
require a move-aware swap, I don't understand why there is a move in
one place and not in the other.

Further, in push(value_type&& x), why write std::move(x)? surely this
should be a no-op?

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All three look ok to me.  Details below:

In article <1183735349.996640.180410@n60g2000hse.googlegroups.com>,
 Chris Jefferson <4zumanga@gmail.com> wrote:

> I've been looking through N2284 and was looking at queue on page 672
> and found some strange inconsistencies.
>
> In operator=(queue&&q),  std::move(q.c) is write.

Right.  The std::move is needed because q.c is an lvalue that needs to
be cast to an rvalue.

> Later in
> swap(queue&&q), the "std::move" is missing. Seeing as most containers
> require a move-aware swap, I don't understand why there is a move in
> one place and not in the other.

In the case of operator= one wants to perform a different operation
depending on whether the rhs is an lvalue or rvalue (one's allowed to
modify the rhs and the other isn't).  In the case of swap, we're saying:
we don't care whether the rhs is an lvalue or rvalue, accept either and
do the same algorithm.

> Further, in push(value_type&& x), why write std::move(x)? surely this
> should be a no-op?

Inside of push(value_type&& x), x is an lvalue.  We want to forward x as
an rvalue to c.push_back, so the std::move is necessary.

-Howard

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