Proto for Boost mailing list page: Re: [proto] restructuring expression

Subject: Re: [proto] restructuring expression
From: Eric Niebler (eric_at_[hidden])
Date: 2012-05-30 21:08:09

On 5/30/2012 4:22 PM, Karsten Ahnert wrote:
> Next trial:
> 
> template< typename I > struct placeholder : I {};
> 
> proto::terminal< placeholder< mpl::size_t< 0 > > >::type const arg1 = {};
> 
> proto::display_expr(
>     fusion::fold(
>         proto::flatten( arg1 ) ,
>         proto::functional::make_terminal()( 1.0 ) ,
>         proto::functional::make_multiplies()
> 	)
>     );
> 
> gives a compilation error:
> 
> /boost/proto/fusion.hpp:86:20: error: no type named ‘proto_tag’ in
> ‘const struct placeholder<mpl_::size_t<0ul> >
>
> It is difficult for me to figure out what happens here. Any ideas?

Right. proto::flatten uses the type of the top-most node to figure out
how to flatten the expression tree. E.g., if you passed arg1 * 32, the
top-most node is a multiplication, so it would produce a list [arg1,
32]. You're passing just a terminal, so it creates a 1-element list
containing the value of the terminal: [placeholder<I>]. Passing this to
proto::functional::make_multiplies results in the above error because
it's expecting a proto expression.

Try this:

    #include <string>
    #include <iostream>
    #include <boost/proto/proto.hpp>

    namespace mpl = boost::mpl;
    namespace proto = boost::proto;
    namespace fusion = boost::fusion;
    using proto::_;

    template< typename I > struct placeholder : I {};

    proto::terminal< placeholder< mpl::size_t< 0 > > >::type const arg1
= {};

    template< class Expr >
    void eval( const Expr &e )
    {
        proto::display_expr(
            fusion::fold(
                proto::flatten( e ) ,
                proto::functional::make_terminal()( 1.0 ) ,
                proto::when<_,
proto::_make_multiplies(proto::_byval(proto::_state), proto::_byval(_))>()
            )
        );
    }

    int main()
    {
        eval( 2 * arg1 * 42.0 * arg1 );
    }

It takes the multiplication tree, flattens it, and turns it back into a
multiplication tree in reversed order.

HTH,

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