Re: [boost] [proto]purpose of binary_expr<T,L,R>::type ?

From: Larry Evans (cppljevans_at_[hidden])
Date: 2008-03-22 18:02:51

On 03/22/08 10:29, Larry Evans wrote:
> On 03/03/08 15:51, Eric Niebler wrote:
> [snip]
>> shift_right<L,R>::type is *very* simple. It is expr<tag::shift_right, 
>> args2<L, R> >. Always. It's just a 2-element container and a tag. If L 
>> and R are expressions, it is an expression. If they're grammars, it is a 
>> grammar, suitable for use as the second template parameter to 
>> proto::matches. It is simple, IMO, and leads to a very straightforward 
> 
> So why is there expr<Tag,Args,Arity>::type if it's always the same
> as its enclosing class, expr<Tag,Args,Arity>.  IIRC, mpl requires it
> for some metaprogramming reason, but it seems that's an implementation
> detail; consequently, although the user can use it, there's no need
> for the user to use it.  If so, then why do many of the examples have:
> 
>    terminal<...>::type a_term_expr={{}};
> 
> instead of just:
> 
>    terminal<...> a_term_expr={{}};
> 
> ?
OOPS, I misread.  I read "shift_right<L,R>::type is shift_right<L,R>"
not "shift_right<L,R>::type is expr<tag::shift_right,args2<L,R> >".
I guess I misread because shift_right<L,R> is a grammar; so, I
assumed that when L and R are grammars, the expr<shift_right,args2<L,R> 
 > grammar would be shirt_right<L,R>.

Sorry.