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

The thing is that y already is a *builtin* function in Haskell. On Fri, 11 Jan 2008 15:59:50 +0200, Achim Schneider wrote:

"Cristian Baboi" wrote:

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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.

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