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Subject: Re: [boost] review request: addition to type_traits library of is_less_comparable<T, U> and others
From: Stewart, Robert (Robert.Stewart_at_[hidden])
Date: 2009-12-08 09:12:54

David Abrahams wrote:
> On Dec 8, 2009, at 6:28 AM, Frédéric Bron wrote:
> > John Maddock wrote:
> >> Rob Stewart wrote:
> >>> Frédéric Bron wrote:
> >>>>> If you really want to support the "don't care"
> >>>>> case for a return type, use has_operator_less<T,
> >>>>> dont_care> where
> >>>>> "dont_care" is a special placeholder type that you define
> >>>>> (could use a better name though?).
> >>>
> >>> I like the idea and the name, though perhaps "any" would
> >>> work, too.
> >>
> >> Doesn't that potentially get confused with boost::any ?

Eh, maybe, but I don't think so. The idea is similar -- any type -- and the documentation will clearly indicate what "any" means in this case. As Frédéric wrote, the test is always for is_convertible_to, not an exact match, so I think the name fits well.

> > It would be has_operator<T, U, boost::type_traits::any>
> > Then boost::type_traits::any_return would be more appropriate.

That is clear, doesn't conflict with boost::any, but also seems overly long. How about "whatever!" :0) Seriously, "anything" might be a good compromise.

> > I prefer void because it is shorter but this could make people think
> > it will check for operator return void... which is not the case; so
> > maybe the long version boost::type_traits::any_return is better.
> By that rationale shouldn't we also require the 3rd parameter
> to be wrapped in is_convertible_to<...> so people don't think
> it's requiring a return type of exactly R (e.g. bool)?

Since is_convertible_to is always implied, and any type can be converted -- using the term loosely -- to no type, "void" might not be confusing. The name used in this context could be left to reviewers to choose.

Rob Stewart robert.stewart_at_[hidden]
Software Engineer, Core Software using std::disclaimer;
Susquehanna International Group, LLP

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