The thing is that y already is a *builtin* function in Haskell.
On Fri, 11 Jan 2008 15:59:50 +0200, Achim Schneider <[EMAIL PROTECTED]> wrote:
"Cristian Baboi" <[EMAIL PROTECTED]> wrote:
>> 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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