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From: Peter Dimov (pdimov_at_[hidden])
Date: 2004-01-23 10:27:31

Douglas Gregor wrote:
> On Thursday 22 January 2004 08:59 am, Peter Dimov wrote:
>> Time will tell. At the moment, it seems to me that ref(x) == ref(y)
>> means &x == &y (at least in bind() context).
> Okay.

So many interesting questions spawned by this technically trivial thing. :-)

Here's another: operator== will introduce yet-another-incompatibility
between boost::bind and boost::lambda::bind. In lambda, operator== creates a
function object. :-)

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