Folks,
I'm having no end of problems with operator<< and proto.
In the beginning, operator << was overloaded for stream output and everything worked OK, then I added the left shift operator<< to the list of protoized operators in the grammar and everything broke. The strange thing is that:
std::cout << a+b;
correctly finds my stream out overload:
template <class Exp> inline std::ostream& operator << (std::ostream& os, const big_number_exp<Exp>& r)
But
std::cout << a;
no longer calls:
template <class Backend> inline std::ostream& operator << (std::ostream& os, const big_number<Backend>& r)
But calls the proto << operator instead.
I'm at a complete loss to explain this, both of my overloads are in the same namespace as their respective types, and as far as I can see both should be preferred to the proto versions.
To answer my own question - in case anyone else gets bitten by this - the answer is you need to provide overloads with both const and non-const reference RHS arguments to effectively hide the proto operator versions. Still some hair left yours.... John.