[Haskell-beginners] Example needed
Sean Perry
shaleh at speakeasy.net
Wed Jun 15 08:21:06 CEST 2011
On Jun 14, 2011, at 8:16 PM, David Place wrote:
>
> On Jun 14, 2011, at 10:01 PM, Zhi-Qiang Lei wrote:
>
>> I wonder why haskell do not take foldl' as default foldl if foldl is not used in production often.
>
> It would be quite interesting to grep over the ghc sources to find uses of foldl.
>
>
From containers/Data/Tree.hs:
-- takes a sequence (queue) of seeds
-- produces a sequence (reversed queue) of trees of the same length
unfoldForestQ :: Monad m => (b -> m (a, [b])) -> Seq b -> m (Seq (Tree a))
unfoldForestQ f aQ = case viewl aQ of
EmptyL -> return empty
a :< aQ' -> do
(b, as) <- f a
tQ <- unfoldForestQ f (Prelude.foldl (|>) aQ' as)
let (tQ', ts) = splitOnto [] as tQ
return (Node b ts <| tQ')
where splitOnto :: [a'] -> [b'] -> Seq a' -> (Seq a', [a'])
splitOnto as [] q = (q, as)
splitOnto as (_:bs) q = case viewr q of
q' :> a -> splitOnto (a:as) bs q'
EmptyR -> error "unfoldForestQ"
haskell2010/Array.hs:
accum :: (Ix a) => (b -> c -> b) -> Array a b -> [(a,c)] -> Array a b
accum f = foldl (\a (i,v) -> a // [(i,f (a!i) v)])
From pretty/Text/PrettyPrint/HughesPJ.hs:
Version 3.0 28 May 1997
* Cured massive performance bug. If you write
foldl <> empty (map (text.show) [1..10000])
you get quadratic behaviour with V2.0. Why? For just the same
reason as you get quadratic behaviour with left-associated (++)
chains.
This is really bad news. One thing a pretty-printer abstraction
should certainly guarantee is insensivity to associativity. It
matters: suddenly GHC's compilation times went up by a factor of
100 when I switched to the new pretty printer.
I fixed it with a bit of a hack (because I wanted to get GHC back
on the road). I added two new constructors to the Doc type, Above
and Beside:
<> = Beside
$$ = Above
Then, where I need to get to a "TextBeside" or "NilAbove" form I
"force" the Doc to squeeze out these suspended calls to Beside and
Above; but in so doing I re-associate. It's quite simple, but I'm
not satisfied that I've done the best possible job. I'll send you
the code if you are interested.
From random/System/Random.hs
{-
If we cannot unravel the StdGen from a string, create
one based on the string given.
-}
stdFromString :: String -> (StdGen, String)
stdFromString s = (mkStdGen num, rest)
where (cs, rest) = splitAt 6 s
num = foldl (\a x -> x + 3 * a) 1 (map ord cs)
time/Data/Time/Format/Parse.hs -- many instances
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