From: Douglas Paul Gregor (gregod_at_[hidden])
Date: 2004-01-23 11:01:58
On Fri, 23 Jan 2004, Peter Dimov wrote:
> 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. :-)
This is fun!
At least we're likely to get a compile-time error if we try to use
operator== in Lambda like we would in Bind. Of course, WWLD? Would Lambda
support some similar form of operator== (perhaps by a
conversion to safe_bool?), or perhaps unlambda()'d function objects would
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