Proto for Boost mailing list page: Re: [proto] Defining the result domain of a proto operator

Subject: Re: [proto] Defining the result domain of a proto operator
From: Eric Niebler (eric_at_[hidden])
Date: 2011-08-26 11:56:56

On 8/26/2011 11:44 AM, Eric Niebler wrote:
> Proto will
> compute the domain of m*v to be matrix. It will use matrix_domain's
> generator to post-process the new expression. That generator can do
> anything -- including placing the new expression in the vector domain.
> In short, there is no requirement that a domain's generator must produce
> expressions in that domain. Just hack matrix_domain's generator.

Expanding on this a bit ... there doesn't seem to a sub-/super-domain
relationship between matrix and vector. Why not make them both (together
with covector) sub-domains of some abstract nt2_domain, which has all
the logic for deciding which sub-domain a particular expression should
be in based on its structure? Its generator could actually be a Proto
algorithm, like:

  nt2_generator
    : proto::or_<
         proto::when<
              vector_grammar
            , proto::generator<vector_expr>(_)>
         proto::when<
              covector_grammar
            , proto::generator<covector_expr>(_)>
         proto::otherwise<
              proto::generator<matrix_expr>(_)>
      >
  {};

  struct nt2_domain
    : proto::domain<nt2_generator>
  {};

Etc...

-- 
Eric Niebler
BoostPro Computing
http://www.boostpro.com