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

From: Geoffrey Romer (geoff.romer_at_[hidden])
Date: 2005-10-06 16:42:13

On 10/6/05, David Abrahams <dave_at_[hidden]> wrote:
> Geoffrey Romer <geoff.romer_at_[hidden]> writes:
>
> > On 10/6/05, David Abrahams <dave_at_[hidden]> wrote:
> >
> >> 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.

Let me put it like this: suppose I have some metaprogramming expression like

typename foo<typename bar<baz>::type >::type

Now suppose I want to apply this operation to a sequence of types,
rather than to baz, using transform<>. To do this, I need to convert
it to a lambda-expression to pass to transform<>. Intuitively, I would
expect this to be done by replacing baz with _1:

typename foo<typename bar<_1>::type >::type

But this doesn't compile for the reasons you explain. You seem to be
telling me to eliminate the typename...::type on metafunction calls
that contain actual placeholders as arguments (and in practice this
seems to work), which leads to:

typename foo<bar<_1> >::type

I take it that's correct? If so, suppose now that I have another piece of code:

typename foo<bar<baz> >::type

In other words, "apply foo to the nullary metafunction bar<baz>". Now
suppose again that I want to turn this into an equivalent
lambda-expression. Intuitively, I would again expect to do this by
substituting _1 for baz:

typename foo<bar<_1> >::type.

However, this is identical to the lambda-expression for my first
example, which has different semantics, so one of the two must be
wrong. Which is wrong, and how can I fix it?

Thanks for all your help, by the way. I'm being a bit stubborn about
this because I want to make sure I really know what's going on here.