On Thursday, September 01, 2011 04:36:46 PM John Maddock wrote:
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:
<snip>
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?
you only validated nullary functions. compare to the call to make_expr: proto::make_expr<proto::tag::function>( detail::sqrt_funct<typename detail::backend_type<Exp>::type>() , arg ) and the proto grammar: proto::function< proto::_ > Which means, you want to match a unary proto expression with a binary proto expression.
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?
Just keep in mind, that proto::function is a expression that can have variadic children. The first is the function expression, and the remaining the function arguments. proto::vararg might help here.
Thanks, John.
_______________________________________________ Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost