Larry Evans wrote:

The doc file:

libs/xpressive/proto/doc/html/boost_proto/user_s_guide/tree_transformations.html

contains:

Second, unary_expr< _, CalculatorGrammar >, has a default transform associated with it. It is a pass-through transform. Given an expression of the form expr< T, arg1< X > >, the transform will invoke the CalculatorGrammar transform (which we haven't completely defined yet -- patience) on X resulting in Y, and then reassemble the expression as expr< T, arg1< Y > >.

yet unary_expr<_, CalculatorGrammar >, defined in traits.hpp, is:

template<typename Tag, typename T> struct unary_expr : has_pass_through_transform<unary_expr<Tag, T> > {...};

so it isn't of the form, expr<T,arg1<X> > for any T or X. OTOH, the nested unary_expr typedef:

typedef expr<Tag, args1<T> > type;

is of this form; so, I guess what's meant by "Given an expression of the form.." is "Given an expression of type, E, where E::type is of the form..."?

What this is trying to say is that unary_expr< _, CalculatorGrammar > IS-A transform, in addition to being a pattern. unary_expr<> also is a meta-function for constructing expr<> types. The three roles are related as follows: 1) unary_expr<T, X>::type is a typedef for expr<T, args1<X> >. 2) unary_expr<T', X'> is a pattern that matches expr<T, args1<X> > IFF T' is T or proto::_, and X' is a pattern that matches X. 3) unary_expr<T', X'>::apply<expr<T, args1<X> >, S, V>::type applies unary_expr<>'s pass-through transform to expr<T, args1<X> > with state S and visitor V. The result is: expr<T, args1< X'::apply<X, S, V>::type> > By "given an expression of the form ...," I mean "when and expression of form ... is passed to the transform's apply<> member template." Sorry for the confusion. I'll make this clearer. -- Eric Niebler Boost Consulting www.boost-consulting.com