Hi Here's another one: f [] = v f (x:xs) = x . f xs The "." is supposed to denote a generic operator - f [] maps to some value v and applied to non-empty list applies head to ? operator. Thanks, Paul
Paul, The Hutton book has gotten great reviews, but pouring over a faulty OCRd version can only be counterproductive. I highly recommend Yet Another Haskell Tutorial[1]. After that you can check out all the other wonderful online resources (starting here [2]). At that point a few OCR errors won't be able to stand in your way! Best wishes on you Haskell explorations, Joel [1] http://www.cs.utah.edu/~hal/docs/daume02yaht.pdf [2] http://haskell.org/haskellwiki/Learning_Haskell
This looks like foldr, specifically:
f = foldr (.) v
so really, there's no way of telling what the operator could be, other
than that:
(.) :: a -> b -> b
where
f :: [a] -> b
v :: b
BTW, I find that most of your emails are in a very large font size. Do
other people notice this? It's not very nice to read. Gmail's 'Show
Original' feature tells me that the HTML version of your message
contains <font size=3> near the start.
On 19/09/2007, PR Stanley
Hi Here's another one: f [] = v f (x:xs) = x . f xs The "." is supposed to denote a generic operator - f [] maps to some value v and applied to non-empty list applies head to ? operator. Thanks, Paul
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as provided, this won't compile because v hasn't been defined.
f [] = 0
f (x:xs) = x . f xs
compiles but gives
f :: (Num (a -> c)) => [c -> c] -> a -> c
which seems wrong. maybe I'm not understanding what v is not supposed to
be.
I am thinking maybe you want iterate
*Main> take 5 $ iterate (+1) 0
[0,1,2,3,4]
It would be easier to understand your desired function if you would
provide an example of desired usage.
t.
PR Stanley
participants (4)
-
Joel Koerwer -
PR Stanley -
Rodrigo Queiro -
Thomas Hartman