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

[David Sabel]

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?

Cheers,
  David