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. :-)