[Haskell-cafe] Re: Why purely in haskell?

[Achim Schneider]

"Cristian Baboi" <cristi at ot.onrc.ro> wrote:

> On Fri, 11 Jan 2008 09:11:52 +0200, Lennart Augustsson  
> <lennart at augustsson.net> wrote:
> 
> > Some people seem to think that == is an equality predicate.
> > This is a big source of confusion for them; until they realize that
> > == is just another function returning Bool they will make claims
> > like [1..]==[1..]
> > having an unnatural result.
> > The == function is only vaguely related to the equality predicate
> > in that it
> > is meant to be a computable approximation of semantic equality (but
> > since it's overloaded it can be anything, of course).
> >
> 
> >> So let's imagine:
> >>
> >> ones = 1 : ones
> >>
> >> ones' = repeat 1
> >>  where repeat n = n : repeat n
> 
> (==) :: Eq a => a -> a -> Bool
> 
> -- what is (y (y) ) by the way ?
> -- how about ( y id ) ?
> 
> y f = f (y f).
> 
> ones :: Num a => [a]
> ones = y (1 :)
> 
> repeat :: a -> [a]
> repeat = \n -> y (n:)
> 
> ones' :: Num a => [a]
> ones' = repeat 1 = (\n->y(n:)) 1 = y (1 : )
> 
> To be able to test them for equality, we must have Eq a.
> 
> So, the reason we cannot test them for equality is that we cannot
> test y (a : ) == y (a : ) where a == a is testable.
> 
Yes, thanks. I actually do think that many things would be easier if
every recursion would be translated to its fixpoint, making the term
tree completely finite and defining y internal, as it's arcane, black
magic.

-- 
(c) this sig last receiving data processing entity. Inspect headers for
past copyright information. All rights reserved. Unauthorised copying,
hiring, renting, public performance and/or broadcasting of this
signature prohibited.