Re: [boost] [Proto] Access to member in proto expression tree

8 Jul 2007


      Replying to myself ...

Eric Niebler wrote:
...
template<typename Expr>
     struct Expression<Expr, typename enable_if<matches<Expr, DoGrammar>
...
::type>
       : extends<Expr, Expression<Expr>, Domain>
     {
         explicit Expression(Expr const &expr = Expr())
           : extends<Expr, Expression<Expr>, Domain>(expr)
           , while_(make_expr<WhileTag, Domain>(This()))
         {}
using extends<Expr, Expression<Expr>, Domain>::operator =;
         typename proto::result_of::make_expr<WhileTag, Domain, 
Expression const>::type while_;
private:
         Expression const &This() const { return *this; }
     };
struct DoWhileGrammar
       : function<
             unary_expr<
                 WhileTag
               , DoGrammar
             >
           , Grammar
         >
     {};
Or perhaps try the following, which is a little cleaner IMO ...


     template<typename Expr>
     struct Expression<Expr
       , typename enable_if<matches<Expr, DoGrammar> >::type
     >
       : extends<Expr, Expression<Expr>, Domain>
     {
         explicit Expression(Expr const &expr = Expr())
           : extends<Expr, Expression<Expr>, Domain>(expr)
         {}

         using extends<Expr, Expression<Expr>, Domain>::operator =;

         template< typename A0 >
         typename proto::result_of::make_expr<
             WhileTag, Domain, Expression const, A0 const
         >::type const
         while_(A0 const &a0) const
         {
             return make_expr<WhileTag, Domain>(*this, a0);
         }
     };

     struct DoWhileGrammar
       : binary_expr<
             WhileTag
           , DoGrammar
           , Grammar
         >
     {};

HTH,

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