Re: [boost] New, powerful way to use enable_if in C++0x

Subject: Re: [boost] New, powerful way to use enable_if in C++0x
From: Stewart, Robert (Robert.Stewart_at_[hidden])
Date: 2011-04-12 09:00:34

Krzysztof Czainski wrote:
> 2011/4/12 Stewart, Robert <Robert.Stewart_at_[hidden]>
> > >
> > I wonder if you couldn't repackage your idea in, say,
> > enable_when and disable_when (plus variants) so that the
> > usage is slightly simpler:
> >
> > class test
> > {
> >   template
> >   <
> >      class... T
> >      , typename boost::enable_when_c<sizeof...(T) == 10>::type = true
> >   >
> >   test(T &&...);
> > };
> >
> > IOW, enable_when_c would expose a nested type, type, of
> > type bool that makes forming the default easy[*].  You could
> > use a pointer or integer type and default to 0, for example,
> > but expressing the default as "= true" just looks good and
> > is easy to remember.
> >
> Enable_when looks good to me, except for one detail:
> "enable_when< [...] >::type = true" may confuse someone, that
> "enable_when< [...] >::type = false" is a synonym for
> "disable_when< [...] >::type = true", so I'm be against the
> part "= true". I'd most like it to be "= 0", but
> "= boost::enabler" looks ok to me too.

I thought about the "= false" issue, but not from the perspective of someone thinking it would negate the logic.  We can document that the convention is to use "= true", but that "= false" has the same effect but with less clarity.  I really doubt anyone would choose "= false" unless they thought it would negate the logic.  Still, I think documentation can prevent that misuse by virtue of a stern warning.

"= 0" would also work, if the type is a pointer or integer type, but other pointer or integer values can then be used, so they really aren't different than true/false.  (Actually, "= 0" would work if the type were bool, too.)  Non-zero values might be thought of as normal, and zero values as negated, logic, thus leading to the same concerns as for bool.

I don't care for boost::enabler because "boost::" and "enable" are repeated (the first "enable" is in "enable_when") and a long line is made even longer.  Note that the use of boost::enabler is optional: the type is void *&, so, for example, "= 0" works just as well and might be thought to negate enable_when.

If enable_when/disable_when were to use a pointer to a UDT and boost::enabler were a pointer of that type, then that might be the only convenient value aside from zero, but that doesn't avoid the possible confusion as with each of the other options.

The only ways I can think of to force one default value is with an enumerated type or a UDT with a member the address of which must be supplied.  The latter requires that the member have external linkage, so the former is simpler:

namespace boost
{
   enum enabler { is_true }; // some clear, short, non-clashing name
}

boost::enable_when<...>::type = boost::is_true

That would use enable_if<...,enabler>::type, of course.

_____
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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