Re: foldr f (head xs) xs is not the same as foldr1 f xs

On Sun, May 08, 2005 at 08:14:30PM +0200, David Sabel wrote:

Hi!

A small example for the claim mentioned in the subject:

Prelude> let x = 1:undefined in foldr (curry fst) (head x) x 1 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... Greetings, Remi -- Nobody can be exactly like me. Even I have trouble doing it.