Re: [boost] [convert] no-throw and fallback feature

Subject: Re: [boost] [convert] no-throw and fallback feature
From: Stewart, Robert (Robert.Stewart_at_[hidden])
Date: 2011-05-11 07:46:44

Gordon Woodhull wrote:
> On May 10, 2011, at 3:55 PM, Vicente BOTET wrote:
>
> I think opt_tie should behave as if it's optional<T>&.  I am
> not sure why one would want to chain assignments in this case,
> but it should be possible.

If you can't explain why it is needed, I consider that a reason to reject it.

Since tie() doesn't work that way, it would be surprising.  However, it would permit Vicente's use case:

int i(fallback);
if (!(opt_tie(i) = convert_to<optional<int>>(s))

> Unfortunately because optional<T&> wants to rebind its
> reference it's not possible to do as boost::tie does.

That is, indeed, unfortunate for this purpose.

> So, it's probably a wrapper.

Agreed.

> This works; I'm not sure if it's optimal:
>
> #include <boost/optional.hpp>
> #include <iostream>
>
> using boost::optional;
> using std::cout;
> using std::endl;
>
> template<typename T>
> class opt_tie_type {
>   T &x_;
> public:
>   opt_tie_type(T &x) : x_(x) {}
>   const optional<T> operator=(optional<T> const& y) {

The return type should be optional<T> const & if not void.

>     if(y)
>       x_ = y.get();
>     return y;
>   }
> };
>
> template<typename T>
> opt_tie_type<T> opt_tie(T &x) {
>   return opt_tie_type<T>(x);
> }

I don't think anything more optimal is possible after addressing the return type of the assignment operator.  However, a robust implementation will need a PTS for reference types to address the reference-to-reference problem (or use remove_reference).

_____
Rob Stewart                           robert.stewart_at_[hidden]
Software Engineer                     using std::disclaimer;
Dev Tools & Components
Susquehanna International Group, LLP  http://www.sig.com

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