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

From: Maurizio Vitale (maurizio.vitale_at_[hidden])
Date: 2007-07-02 10:14:20

On Jul 1, 2007, at 6:24 PM, Eric Niebler wrote:

>
> But now that I think about it some more, Dan probably wants if_ to  
> be an
> actual function instead of a Proto terminal, because phoenix  
> expressions
> have an operator() that does something special. You can use the
> BOOST_PROTO_DEFINE_FUNCTION_TEMPLATE macro for this.
>
>    BOOST_PROTO_DEFINE_FUNCTION_TEMPLATE(
>        1
>      , if_
>      , phoenix::phoenix_domain
>      , (phoenix::tag::if_)
>      , BOOST_PP_SEQ_NIL
>    )
>
>
Eric, what is the meaning of the last argument (empty here and called  
BOUNDARGS in the sources)?

The only example I could find is in mixed.cpp where it seems to be  
used to allow the tag to be unrelated to
the function being defined. Following that example:
        BOOST_PROTO_DEFINE_FUNCTION_TEMPLATE(
                1,
                if_,
                phoenix::phoenix_domain,
                proto::tag::function
                ((if_struct))
        )
would also work and create a proto::expr with a tag::function as tag  
and first argument a (terminal?) if_struct.
It assumes if_struct has a default constructor.
Is this the right interpretation?

In the mixed.cpp example is defining struct sin_ { } the way it is  
enough to allow the default_context to evaluate
sin_(expr)?