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Not necessarily. These these can be added to your DSEL post-hoc, and there are some fun games you can play with proto's operator->*. Consider the following program:
Thanks, Eric. That was an Ah-ha moment for me. I tried a different variation of your program. I changed point to a templated tuple<typename X, typename Y> and declared the terminals x and y with proto::_ template arguments. Here is the complete program : #include <cassert> #include <boost/proto/proto.hpp> #include <iostream> namespace proto = boost::proto; template<typename X, typename Y> struct tuple { X x; Y y; }; template<typename X, typename Y> struct tuple_types { typedef X tuple<X, Y>::* X_type; typedef Y tuple<X, Y>::* Y_type; }; //Does not work // proto::terminal<tuple_types<proto::_, proto::_>::X_type>::type x // = {&tuple<proto::_, proto::_>::x}; // proto::terminal<tuple_types<proto::_, proto::_>::Y_type>::type y // = {&tuple<proto::_, proto::_>::y}; proto::terminal<tuple_types<int, float>::X_type>::type x = {&tuple<int, float>::x}; proto::terminal<tuple_types<int, float>::Y_type>::type y = {&tuple<int, float>::y}; struct arg_ {}; proto::terminal<arg_>::type arg = {{}}; struct micro_lambda : proto::or_< proto::when< proto::terminal<arg_> , proto::_state
, proto::otherwise< proto::_default<micro_lambda>
{} eval; using namespace std; int main() { tuple<int, float> t; t.x = 1; t.y = 41.f; cout << eval(arg->*x + arg->*y, t) << endl; } When I declare x and y like shown in the commented out part, I get compile errors. What am I doing wrong there? Do I need extra grammar rules to match the member pointers of a templated class? Manjunath