Re: [boost] [boost:proto] adding new tags

From: Eric Niebler (eric_at_[hidden])
Date: 2007-05-25 19:18:59

Maurizio Vitale wrote:
> Suppose I needed a new tag for representing some high-level concept that I do not want
> to express in terms of C++ operators [it could be a get_bit(N) operation that has very
> different implementation when for my numbers the unserlyining implementation is a
> builtin type or a GMP big int]
> 
> Here's my code for adding a binary operator my_plus:
> 
>     struct my_plus {};
> 
>     template<typename Left, typename Right>
>     proto::expr<my_plus, proto::args2<proto::ref_<Left>, proto::ref_<Right> > > const
>     make_my_plus_expr(Left& left, Right& right)
>     {
>       typedef proto::expr<my_plus, proto::args2<proto::ref_<Left>, proto::ref_<Right> > > expr_type;
>       expr_type that={left,right};
>       return proto::generate<typename Left::domain, expr_type>::make (that);
>     }
> 
> Questions:
> 
>   - is the above the right way, or is there some friendlier interface?

You can now do this with make_expr. It would look like this:

   struct my_plus {};
   proto::functional::make_expr<my_plus> const make_my_plus_expr = {};

Then you can say make_my_plus_expr(e1, e2) and get a binary expr with 
my_plus as a tag type. If either e1 or e2 are not yet a proto 
expression, they are made into proto terminals first.

If you want to specify a domain, so the resulting expr is wrapped in 
some extension wrapper, you can specify the domain as the second 
template parameter, as

   struct my_plus {};
   proto::functional::make_expr<my_plus, my_domain> const
       make_my_plus_expr = {};

Currently, the domains of the children expressions are *not* considered 
in either case. This seemed simpler. Let me know if this doesn't meet 
you needs. I can imagine using a pseudo-domain for the second parameter 
that means: propagate the domain from the children nodes.

-- 
Eric Niebler
Boost Consulting
www.boost-consulting.com