Re: [Boost-users] [Proto]: How to use complex grammars in a domain/extension

Subject: Re: [Boost-users] [Proto]: How to use complex grammars in a domain/extension
From: Eric Niebler (eric_at_[hidden])
Date: 2010-09-21 09:43:06

On 9/21/2010 5:55 AM, Thomas Heller wrote:
> Solved the mistery. here is the code, explanation comes afterward:
> 
> #include<boost/proto/proto.hpp>
> 
> using namespace boost;
> 
> typedef proto::terminal<int>::type terminal;
> 
> struct addition:
>     proto::or_
>     <
>        proto::plus<addition, addition>,
>        proto::terminal<proto::_>
>     >
> {};
> 
> struct equation:
>     proto::or_
>     <
>         proto::equal_to<addition, addition>
>     >
> {};
> 
> template<class Expr>
> struct extension;
> 
> struct my_domain:
>     proto::domain
>     <
>          proto::pod_generator<  extension>,
>          // we need both grammars in our domain grammar
>          proto::or_<equation, addition>,

That will also work because addition (contrary to its name) will also
match lone terminals.

>          proto::default_domain
>     >
> {};
> 
> 
> template<class Expr>
> struct extension
> {
>      BOOST_PROTO_EXTENDS(
>          Expr
>        , extension<Expr>
>        , my_domain
>      )
> };
> 
> template <typename Grammar, typename Expr>
> void matches(Expr const&)
> {
>      std::cout << std::boolalpha
>          << proto::matches<Expr, Grammar>::value << "\n";
> }
> 
> int main()
> {
>      extension<terminal>  i;
>      extension<terminal>  j;
> 
> 
>      matches<equation>(i);              // 1) false
>      matches<equation>(i == j);         // 2) true
>      matches<equation>(i == i + i);     // 3) true
>      matches<equation>(i + i == i);     // 4) true
>      matches<equation>(i + i == i + i); // 5) true
>      matches<equation>(i + i);          // 6) false
> }
> 
> Ok, what happened, why does this work, and why didn't the previous attempt 
> work:
> 
> the equation grammar obviously doesn't match case 1) and 6). And that is 
> exactly why proto didn't generate any operator+ overload.
> In order for a proto overload to be created the following conditions must be 
> true:
> 
> 1) the operands must be in a compatible domain
> 2) the left hand operand and the right hand operand must match the grammar 
> specified in the domain

Precisely.

> 3) the resulting expression must match the grammar specified in the domain.

Right-o.

> In our example 1) is always true. 2) and 3) are not true for expressions 
> like 1) and 6). That means, that any binary operator gets disabled whenever 
> the LHS or the RHS is a terminal<_>. because of the equation grammar, 
> terminals or additions can not stand alone, that is why operator creation 
> failed. Makes sense!
> In order to enable the operators, we just say, our domain can have addition 
> _or_ equation. That's it, operator+ and operator== get enabled.
> We can still detect invalid expression with proto::matches!
> 
> So, everything works as expected!

Thomas wins the prize. This is not a bug in Proto.

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