On 8/24/2011 2:07 PM, John Maddock wrote:

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 guessing that big_number<T> inherits from big_number_exp<proto::terminal<T>::type>. Did I guess right? I'm also guessing that in "std::cout << a", that a is a non-const big_number. Did I guess right?

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.

Before I start tearing my hair out, anyone any ideas?

Yes. It's picking proto's right-shift operator because "a" is non-const. It's preferred because your operator<< requires a cv conversion. Aside: "a+b" returns a const-qualified rvalue, assuming it's using Proto's operator+. So no problem there. So, one solution would be to just add a stream inserter (and extractor, if desired) for non-const big_numbers. You'll still have problems, though. Try: std::stringstream str; str << a; Now, the LHS isn't *exactly* std::ostream&, and you're back where you started. Guessing. If that's the case, let me know and I'll help you modify your grammar appropriately. -- Eric Niebler BoostPro Computing http://www.boostpro.com