foldr f (head xs) xs is not the same as foldr1 f xs
rturk at science.uva.nl
Sun May 8 14:34:26 EDT 2005
On Sun, May 08, 2005 at 08:14:30PM +0200, David Sabel wrote:
> A small example for the claim mentioned in the subject:
> Prelude> let x = 1:undefined in foldr (curry fst) (head x) x
> Prelude> let x = 1:undefined in foldr1 (curry fst) x
> *** Exception: Prelude.undefined
> Perhaps it would be better to change the implementation of foldr1?
Why? *wonders what he's missing* It sounds like a rather silly
claim to me. When changed to
foldr f (head xs) (tail xs) is not the same as foldr1 f xs
I would be more interested to see examples...
Nobody can be exactly like me. Even I have trouble doing it.
More information about the Glasgow-haskell-users