Re: [Boost-users] [Proto]: How to use complex grammars in a domain/extension

On 09/21/2010 11:55 AM, Thomas Heller wrote:

<snip> Solved the mistery. here is the code, explanation comes afterward:

<snip> So, everything works as expected!

Thomas, wow, this was driving me crazy, thanks for solving and explaining in all the detail! :-) I am still not sure if this isn't a conceptual problem, though: The "equation" grammar is perfectly OK. But without or-ing it with the "addition" grammar, the extension does not allow ANY valid expression to be created. I wonder if that is a bug or a feature? Thanks and regards, Roland

#include

using namespace boost;

typedef proto::terminal<int>::type terminal;

struct addition: proto::or_ < proto::plus, proto::terminalproto::_ > {};

struct equation: proto::or_ < proto::equal_to > {};

template<class Expr> struct extension;

struct my_domain: proto::domain < proto::pod_generator< extension>, // we need both grammars in our domain grammar proto::or_, proto::default_domain > {};

template<class Expr> struct extension { BOOST_PROTO_EXTENDS( Expr , extension<Expr> , my_domain ) };

template void matches(Expr const&) { std::cout<< std::boolalpha << proto::matches::value<< "\n"; }

int main() { extension<terminal> i; extension<terminal> j;

matches<equation>(i); // 1) false matches<equation>(i == j); // 2) true matches<equation>(i == i + i); // 3) true matches<equation>(i + i == i); // 4) true matches<equation>(i + i == i + i); // 5) true matches<equation>(i + i); // 6) false }