Re: [boost] On bind::operator==, again

From: Peter Dimov (pdimov_at_[hidden])
Date: 2004-03-11 06:14:50

Douglas Gregor wrote:
>
> We can have boost::function_equal(f, g), but I'd prefer that it
> always use operator==. Then boost::lambda, boost::bind, etc. can just
> overload function_equal.

That's what I had in mind when I posted; sorry for not mentioning it, but I
didn't want to influence the answers. Am I not evil.

In summary, unqualified function_equal(f, g), with a default implementation
that returns f == g.

> Why? Because most users just want to write a
> normal operator== and have it work; that's what makes sense. Just
> because Lambda can do some screwy things with syntax (and believe me,
> I love Lambda) doesn't mean others should have to use a weird
> comparison syntax.

I note that you show some pro-== bias. function_equal (my original name was
fn_equal, but I can live with either) is not the same as ==, even though it
does use == by default. Placeholders, reference_wrapper, tuple all respond
differently to fn_equal and ==.

In fact, I still think that your decision to expose the 'f contains g'
operation under the f == g notation is a mistake, too. 'contains' is not
'==', either; it is asymmetric, as it is possible for f to contain g but not
vice versa.

In short, == for function objects is an illusion and it must die, Herb
Sutter's influential articles notwithstanding.