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

From: Larry Evans (cppljevans_at_[hidden])
Date: 2008-03-21 22:48:23

On 03/04/08 11:38, Eric Niebler wrote:
> Larry Evans wrote:
>> On 03/03/08 19:29, Eric Niebler wrote:
>>  > Larry Evans wrote:
>>  >>      If T0,T1,...,Tn are all expression types, and Tag is an n-ary tag,
>>  >>      then expr<Tag,T0,T,...,Tn> is an expression type.
>>  >
>>  > Yes. It is also a grammar type that matches itself.
>>  >
>> [snip]
>> Hmm...  So a grammar is an expression and an expression is a grammar.
>> Sorta like lisp where a program is data and data is a program.
>> I guess proto::matches is like lisp eval.  Is that sorta right?
> 
> An expression is a grammar, but a grammar is not an expression. For 
> example, not_<X> is only a grammar and not an expression.
> 

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>)
   ;

gram //describes a grammar
   = expr
   | wildcard //struct wildcardns_::_ in matches.hpp
   | control  //any struct or class template in namespace control in
              //matches.hpp
   ;
control
   = not_ >> gram
   | or_  >> gram >> gram+
   | and_ >> gram >> gram+
   | if_  >> gram >> gram
   ;

The  '= expr' as first alternate of gram reflects the statement
'an expression is a grammar'.  The absence of gram on the
rhs of the expr equation reflects the statement 'a grammar
is not an expression'.

Is that about right?