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

From: Larry Evans (cppljevans_at_[hidden])
Date: 2008-03-04 09:35:09

On 03/03/08 19:29, Eric Niebler wrote:
 > Larry Evans wrote:
[snip]
 >> was formed with the same template as an expression... Hmm..  Getting
 >> confused again. How can one distinguish a grammar from an expression
 >> if both are formed from the same template, i.e. the expr template?
 >> Maybe by starting from the base case, the terminal template?
 >> IOW, is the following:
 >>
 >>    base_case)
 >>
 >>      For any type T:
 >>        terminal<T> is a grammar type.
[snip]
 >>        terminal<T>::type is an expression type.
 >
 > Yes. It's also a grammar type. It matches itself (more or less -- see
 > the reference for proto::matches<>).
 >
 >>    inductive_case)
 >>
 >>      If T0,T1,...,Tn are all grammar types, and Tag is an n-ary tag,
 >>      then expr<Tag,T0,T2,...,Tn> is a grammar type.
 >
 > Yes. (Minor correction, expr<Tag, argsN<T0,T1,...,Tn> > is a grammar
 > type.)
[snip]
 >>      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?