Re: [Boost-users] [mpl] what is not a placeholder expression

From: Joaquin M Lopez Munoz (joaquin_at_[hidden])
Date: 2007-07-28 03:24:45

David Abrahams <dave <at> boost-consulting.com> writes:

> on Fri Jul 27 2007, "JOAQUIN LOPEZ MU?Z" <joaquin-AT-tid.es> wrote:
> 
> > According to the MPL reference, a placeholder expression
> > X<a1,..,an> must satisfy (among others) the following
> > condition:
> >
> >   "All of X's template parameters, including the
> >   default ones, are types."
> >
> > According to that the following should then compile
> > without errors:
[...] 
> I don't see how you can conclude the above will compile without error
> just from (a part of) the definition of placeholder expression.
> 
> Are you saying that since foo<_1> is not a placeholder expression,
> when applied to int it should be treated as a metafunction class?

Yes, this is exactly my point. Another way in which foo<_1>
can be forced to not be treated as a ph expression is this:

  // no of parms > BOOST_MPL_LIMIT_METAFUNCTION_ARITY
  template<typename T,
    typename=int,typename=int,typename=int,
    typename=int,typename=int> 
  struct foo 
  {
    template<typename Q>
    struct apply{
      typedef Q type;
    };
  };

You can try yourself this variation and see that the program
really resolves to t1 <- int. But for some reason, the
int=0 variation does not do the same, at least in GCC 3.4.4.

Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo

PS: I'd greatly appreciate if you could you try the snippet
(the int=0 version) with other compilers --my current compiler
count is so very limited. Thx!

PPS: In case you're wondering, this is not a purely academic
question: I've got a real world scenario where I want to
prevent foo<x> from being treated as a ph expression, namely
when foo<x> is a metafunction class parameterized by
a lambda expression (i.e. a ph expression or metafunction
class) x.