I believe R J was consciously restricting himself to finite lists in the
original post.
2009/3/18 MigMit
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