[boost] [proto] Can't use functions in expressions?

Subject: [boost] [proto] Can't use functions in expressions?
From: John Maddock (boost.regex_at_[hidden])
Date: 2011-09-01 11:36:46

So.... I have a function that returns an expression template of 
proto::tag::function type, like so:

template <class Exp>\
typename proto::result_of::make_expr<\
    proto::tag::function\
  , detail::BOOST_JOIN(func, _funct)<typename 
detail::backend_type<Exp>::type>\
  , detail::big_number_exp<Exp>\
>::type const \
func(const detail::big_number_exp<Exp>& arg)\
{\
    return proto::make_expr<proto::tag::function>(\
        detail::BOOST_JOIN(func, _funct)<typename 
detail::backend_type<Exp>::type>() \
      , arg   \
    );\
}\

and it all works fine, until I try and use the result of such a function 
call in an expression:

sqrt(arg) + arg

yeilds:

1>m:\data\boost\sandbox\big_number\libs\math\ide\extended_real\scrap\scrap.cpp(25): 
error C2893: Failed to specialize function template 'const 
boost::proto::detail::enable_binary<boost::proto::domainns_::deduce_domain,boost::proto::detail::not_a_grammar,boost::mpl::or_<boost::proto::is_extension<T>,boost::proto::is_extension<Right>>,boost::proto::tag::plus,const 
Left,const Right>::type boost::proto::exprns_::operator +(const Left &,const 
Right &)'
1>          With the following template arguments:
1>          'boost::math::detail::big_number_exp<Expr>'
1>          with
1>          [
1> 
Expr=boost::proto::exprns_::basic_expr<boost::proto::tag::function,boost::proto::argsns_::list2<boost::math::detail::big_number_exp<boost::proto::exprns_::basic_expr<boost::proto::tag::terminal,boost::proto::argsns_::term<boost::math::detail::sqrt_funct<boost::math::mpfr_real_backend<50>>>,0>>,boost::math::detail::big_number_exp<boost::proto::exprns_::expr<boost::proto::tag::terminal,boost::proto::argsns_::term<boost::math::big_number<boost::math::mpfr_real_backend<50>> 
*>,0>>>,2>
1>          ]
1>          'boost::math::mpfr_real_50'

So I figured I needed to update my grammar, so I tried adding this case:

template<>
struct big_number_grammar_cases::case_<proto::tag::function>
  : proto::function< proto::_ >
{};

But still no joy (same error message).

Any ideas on what I'm doing wrong?

BTW I couldn't find any examples of using proto::function in a grammar, it 
wasn't obvious to me what it's template arguments should be, or indeed 
whether this actually enabled the function call operator rather than my use 
case?

Thanks, John.