More interesting:
foldl (flip const) whatever (repeat 1 ++ [1,2,3])
Daniel Fischer wrote on 18.03.2009 15:17:
Am Mittwoch, 18. März 2009 13:10 schrieb Daniel Fischer:_______________________________________________
Now prove the
Lemma:
foldl g e (ys ++ zs) = foldl g (foldl g e ys) zs
for all g, e, ys and zs of interest.
(I don't see immediately under which conditions this identity could break,
maybe there aren't any)
Of course, hit send and you immediately think of
foldl (flip const) whatever (undefined ++ [1,2,3])
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe