On Wed, Aug 18, 2010 at 4:38 PM, Lennart Augustsson <
lennart.augusts...@gmail.com> wrote:
> You don't know that f is strict in its first argument so you cannot
> deduce that go is strict in z in the first case.
>
I'm not sure I understand.
f :: Int -> Int -> Int -> Int
f = \x y z -> x + y + z
On Wed, Aug 18, 2010 at 2:47 PM, Ian Lynagh wrote:
> On Wed, Aug 18, 2010 at 12:01:54PM +0200, Johan Tibell wrote:
> >
> > foldlWithKey' :: (b -> k -> a -> b) -> b -> Map k a -> b
> > foldlWithKey' f z0 m = go z0 m
> > where
> > go z Tip = z
> > go z (Bi
On Wed, Aug 18, 2010 at 12:01:54PM +0200, Johan Tibell wrote:
>
> foldlWithKey' :: (b -> k -> a -> b) -> b -> Map k a -> b
> foldlWithKey' f z0 m = go z0 m
> where
> go z Tip = z
> go z (Bin _ kx x l r) = let x' = f (go z l) kx x in x' `seq` go x' r
>
> C
Hi,
I was adding a strict pre-order fold to the Data.Map module and I ran into
this slightly surprising behavior. Modeled on foldl' for lists I defined
foldlWithKey' :: (b -> k -> a -> b) -> b -> Map k a -> b
foldlWithKey' f z0 m = go z0 m
where
go z Tip = z
