Kim Kuen Tang wrote:
Hi all,
let _o_ be any proto terminal and v a vector. I would like to extract the vector v from the expression _o_(v) to do some modification on this vector. For this purpose i defined a grammar Rhs to extract the most right hand side of the tree. But unfortunately the compiler is asking me for constness for this code:
boost::result_of<Rhs(Expr)>::type & result = Rhs()(expr);
Whereas the expression _o_ + v compiles fine without asking me for constness.
Am I missing something?
You're not missing anything. Proto handle operator() differently than the other operators: operator() captures its arguments by const ref, whereas the other operators have overloads for const and non-const ref. Why the difference? Because operator() can have up to BOOST_PROTO_MAX_FUNCTION_CALL_ARITY arguments, which defaults to 5. To accomodate every permutation of const and non-const arguments would require 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 overloads, which is ... a lot. So what are your options. my suggestion is to first change this: boost::result_of<Rhs(Expr)>::type & result = Rhs()(expr); to more correct: boost::result_of<Rhs(Expr&)>::type result = Rhs()(expr); Now, if you are certain that the result is really a mutable lvalue, you can const_cast the result to a std::vector<int>& and mutate it. The alternative would be to make _o_ an expression extension with overloads of operator() to handle all the permutations. -- Eric Niebler BoostPro Computing http://www.boostpro.com