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From: Markus Schöpflin (markus.schoepflin_at_[hidden])
Date: 2002-06-12 11:22:57

Peter Dimov wrote:
> From: "Markus Schöpflin" <markus.schoepflin_at_[hidden]>


> > std::scoped_ptr<Obj> p(create_obj());
> No, unfortunately it will not. The constructor will take a non-const
> reference to auto_ptr, and those cannot be bound to rvalues. std::auto_ptr
> uses an interesting (and obscure) technique to allow

Hmm, I just hacked my copy of scoped_ptr.hpp and added

explicit scoped_ptr(std::auto_ptr<T> &p): ptr(p.release()) {}

and it works as expected, taking away ownership from auto_ptr and giving
it to scoped_ptr.

Is this something my compiler (MSVC6SP5) allows as an extenstion?

> std::auto_ptr<Obj> p(create_obj());
> and I'm not sure that we need to enable this functionality for scoped_ptr.

Well, it would make the intention clear that I want to take ownership
of the generated object and keep it.

OTOH, it works if you write

std::scoped_ptr<Obj> p(create_obj().release());

but it looks su much nicer if you can ommit the release()... :-)


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