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

From: Douglas Gregor (gregod_at_[hidden])
Date: 2004-03-10 00:09:34

On Tuesday 09 March 2004 10:00 am, Peter Dimov wrote:
> What do you think this expression should do:
>
>     bind(f, _1) == bind(g, _1)
>
> Currently, boost::bind returns f == g when f and g are of the same type,
> and fails at compile type otherwise (this is the equality comparison
> feature Doug requested).
>
> lambda::bind returns lambda(x) { return f(x) == g(x) }.

Well, the lambda function object returned by bind(f, _1) == bind(g, _1) could 
be convertible to bool. Too cute? It might just be slick enough to work.

        Doug