[Haskell-cafe] What class for splittable data / balanced-fold?

Ryan Newton rrnewton at gmail.com
Mon Sep 30 03:13:06 CEST 2013


Thanks, that's interesting to know (re: Fortress).

Interestingly, in my Fortress days we looked at both using a split-like
> interface and at a more foldMap / reduce - like interface, and it seemed
> like the latter worked better – it requires a lot less boilerplate for
> controlling recursion, and better matches the fanout of whatever structure
> you're actually using underneath.
>

Ok, we'll have to try that.  I may be underestimating the power of a
newtype and a monoid instance to expose the structure..  I was wrong about
this before [1].  Here's the foldMap instance for Data.Map:

  foldMap _ Tip = mempty  foldMap f (Bin _ _ v l r) = Foldable.foldMap
f l `mappend` f v `mappend` Foldable.foldMap f r

Simon Marlow in his recent Haxl talk also had a domain where they
wanted a symmetric (non-monadic) parallel spawn operation...

But it remains pretty hard for me to reason about the operational
behavior of these things... especially since foldMap instances may
vary.

Thanks,

   -Ryan

[1] For example, here is a non-allocating traverseWithKey_ that I
failed to come up with:


-- Version of traverseWithKey_ from Shachaf Ben-Kiki
-- (See thread on Haskell-cafe.)
-- Avoids O(N) allocation when traversing for side-effect.

newtype Traverse_ f = Traverse_ { runTraverse_ :: f () }
instance Applicative f => Monoid (Traverse_ f) where
  mempty = Traverse_ (pure ())
  Traverse_ a `mappend` Traverse_ b = Traverse_ (a *> b)
-- Since the Applicative used is Const (newtype Const m a = Const m), the
-- structure is never built up.
--(b) You can derive traverseWithKey_ from foldMapWithKey, e.g. as follows:
traverseWithKey_ :: Applicative f => (k -> a -> f ()) -> M.Map k a -> f ()
traverseWithKey_ f = runTraverse_ .
                     foldMapWithKey (\k x -> Traverse_ (void (f k x)))
foldMapWithKey :: Monoid r => (k -> a -> r) -> M.Map k a -> r
foldMapWithKey f = getConst . M.traverseWithKey (\k x -> Const (f k x))
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