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

From: Larry Evans (cppljevans_at_[hidden])
Date: 2008-03-24 12:10:08

On 03/23/08 16:02, Larry Evans wrote:
> On 03/22/08 22:12, Eric Niebler wrote:
>  > Larry Evans wrote:
> [snip]
>  >> The following grammar I think summarizes the difference between
>  >> grammar and expression:
>  >>
>  >> expr //describes an expression.
>  >>    = terminal
>  >>    | expr+
>  >>      //i.e. 1 or more expressions (e.g. expr<tag::plus,expr0,expr1>)
>  >>    ;
> 
> In addition, I wanted some way to tell what's
> a valid 1st arg to matches.  The above description of expr is for a
> grammar.  Now a valid 1st arg to matches would apply ::type to each
> node in the above tree, AFAICT.  At first, I thought of:
> 
>    expr_type
>       = terminal::type
>       | expr_type+
>       ;
> 
> but that would make:
> 
>       ( terminal::type, terminal::type, terminal::type )
> 
> a valid expresion when what's needed is:
> 
>       ( terminal::type, terminal::type, terminal::type )::type
> 

OOPS (again).  I guess this would produce what's needed:

   expr_type
      = terminal::type
      | (expr_type+)::type
      ;