Re: [boost] [proto] Problems with operator <<

Subject: Re: [boost] [proto] Problems with operator <<
From: Eric Niebler (eric_at_[hidden])
Date: 2011-08-24 15:37:12

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