Sorry for the delay. I was away for a bit. On 2/18/2010 1:12 PM, Manjunath Kudlur wrote:

I am trying understand the use of proto::fold to handle vararg expressions. As an experiment, I am trying to convert expressions of the form foo(a,b,c) to a string of the form "foo(a,b,c)". I currently have the following code where I treat the vararg expression as a fusion sequence and convert each parameter to a string using appropriate grammar.

More or less OK so far. In the operator() of _foo_tostr, I have a case

for each length of the vararg expression. Please let me know if there are better ways of doing this.

<snip> There's room for improvement here. Like you, I found it simpler to treat the vararg as a Fusion sequence than trying to use proto::fold on the expression. But your code can be made simpler with fusion::fold. See below. #include <string> #include <sstream> #include <iostream> #include #include using namespace std; namespace proto = boost::proto; namespace fusion = boost::fusion; using proto::_; unsigned ids; struct var { unsigned id; var() { id = ++ids; } }; struct foo {}; typedef proto::terminal<var>::type Var; typedef proto::terminal<foo>::type Foo; string _var_tostr(string const &str, Var const &v) { stringstream sout; if(!str.empty()) sout << str << ", "; sout << "v" << proto::value(v).id; return sout.str(); } struct _foo_tostr : proto::callable { typedef string result_type; template<typename T> result_type operator()(T const &e) const { return "foo(" + fusion::fold(e, string(), &_var_tostr) + ")"; } }; struct foo_tostr : proto::when< proto::function< proto::terminal<foo>, proto::vararg >, // Pop the foo terminal off the front of the // current expression (which is a valid Fusion // sequence) and pass the result to _foo_tostr. _foo_tostr(proto::functional::pop_front(_)) > {}; int main() { Var a, b, c; Foo f; cout << foo_tostr()(f(a,b,c)) << endl; } HTH, -- Eric Niebler BoostPro Computing http://www.boostpro.com