Re: [boost] review request: addition to type_traits library of is_less_comparable<T, U> and others

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: 2009-12-07 13:49:50

Christian Schladetsch wrote:
> On Mon, Dec 7, 2009 at 9:12 PM, John Maddock
> <john_at_[hidden]> wrote:
>
> > operator    name in <functional>    proposed name (existence)
> >> proposed name (existence and non void return type convertible to
> >> something)
> >> <           less                    has_operator_less
> >> operator_less_has_standard_behaviour

If there were reason to keep it, I'd suggest something more like "std_operator_less," but that suggests a Standard-mandated interface, which there isn't (for UDTs).

> > I don't like operator_less_has_standard_behaviour at all.
> > In fact is there really any point to it?
> >
> > Why not simply:
> >
> > has_operator_less<T, R = bool>

has_operator_less<T, U, R = bool> is nice.

> > Where R is the expected return type.  In all but a
> > vanishingly small number of cases R will be bool.

Agreed.

> > 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.

> Perhaps
>
> boost::has_operator_less<T,R=bool>
>
> Is "less ugly" than
>
> boost::operators::less::exists<T, R=bool>
>
> But is it 'better'?

It is more in keeping with the existing traits, so if the latter were offered, the former would still be appropriate.

> Using boost::operators::[name] has enough potential benefits
> that I don't see why it should be rejected so casually:
>
> * boost::operators::[tab]::[tab] works better with Intellisense

I'm sure that holds sway with many, but I don't care about it.

> * it is factored over the operators, which provides a common
> structure for all operators in a tree

The structure is nice, but "has_operator_<op>" with a return type template argument also provides structure.

> * it provides a single type that designates each operation
> uniquely. that is, operators can become first-class objects.

There may be significant value on this point.

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

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