Re: [Boost-users] [mpl] trouble with conditionals

From: David Abrahams (dave_at_[hidden])
Date: 2005-10-06 16:08:55

Geoffrey Romer <geoff.romer_at_[hidden]> writes:

> On 10/6/05, David Abrahams <dave_at_[hidden]> wrote:
>> Geoffrey Romer <geoff.romer_at_[hidden]> writes:
>>
>> >> > If so, doesn't that mean Cromwell's fix is incorrect?
>> >>
>> >> I'm afraid I've not seen his fix.
>> >
>> I'm also afraid I don't have time to look at it right now.
>
> Fair enough. It's sort of a side issue anyway.
>
>> > What I don't
>> > understand is when I should be saying "typename
>> > metafunc<args>::type" and when I should be using plain
>> > "metafunc<args>". This isn't a language-correctness issue, it's an
>> > MPL usage issue.
>>
>> Oh.  Maybe mapping into the runtime world will help.  If you think of
>> metafunc as a regular function, then "metafunc<args...>::type" is
>> equivalent to a regular function call, and "metafunc<args...>" is
>> equivalent to boost::bind(func, args...)
>>
>> These are exact analogies, AFAICT.
>
> OK, that's what I thought, but it seems not to apply to
> lambda-expressions. If I want to call foo<> on a placeholder

I don't know what you mean, but I'm pretty sure you don't mean what
you said.  Nobody but the MPL internals want to actually invoke
metafunctions on placeholders.  Very often that will cause an error,
because the placeholder does not meet the expectations of the
metafunction for the type that is actually supposed to be passed to
it.

> I seem to be required to write foo<_1>, and not typename
> foo<_1>::type, even though I want foo<> to actually be applied to
> the placeholder (after substitution).  What am I missing?

Well, it's the "after substitution" bit.  For the reasons described
above, you can't in general instantiate foo<_1>, and even if you did,
the result might be int, e.g.

    template <class T>
    struct foo
      : mpl::if_<boost::is_pointer<T>, T, int>
    {}

    // foo<_1> => int

and there's no information left in the type int that can tell the
library where to make a substitution, or even what template into which
it should be substituted.

So, in placeholder expressions, you want to delay evaluation, and you
don't actually add the ::type.

HTH,

-- 
Dave Abrahams
Boost Consulting
www.boost-consulting.com