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

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