On Sat, Oct 20, 2001 at 01:11:05AM +1000, Andrew J Bromage wrote:
> G'day all.
>
> On Fri, Oct 19, 2001 at 02:30:59PM +0100, Ian Lynagh wrote:
>
> > Also, the prelude definition of zipWith has LVL whereas the following
> > definition has LVV. Why is something like the following not used?
> >
> > > zipWith :: (a->b->c) -> [a] -> [b] -> [c]
> > > zipWith f (a:as) (b:bs) = f a b : zipWith f as bs
> > > zipWith _ _ [] = []
> > > zipWith _ _ _ = []
>
> Generally speaking, Haskell programmers don't like inserting more code
> with the only effect being to make the function less lazy. This goes
> double for standard library code.
>
> I say "generally" because occasionally there's a good reason (e.g.
> forcing evaluation can make a program more space-efficient). Is there
> a good reason behind your version of zipWith other than the strictness
> signature being more symmetrical? :-)
With the following code:
main :: IO()
main = putStrLn $ show $ last $ zipa list (cycle "hello world")
where list = concat $ replicate 1000000000 $ replicate 10 'a'
zipa :: [a] -> [b] -> [(a, b)]
zipa (x:xs) (y:ys) = (x, y):zipa xs ys
zipa _ _ = []
zipb :: [a] -> [b] -> [(a, b)]
zipb (x:xs) (y:ys) = (x, y):zipb xs ys
zipb _ [] = []
zipb _ _ = []
I get
$ time nice -n 10 ./W
('a','h')
real 148m50.013s
user 146m37.930s
sys 0m4.470s
$
whereas changing the zipa to zipb gives me
$ time nice -n 10 ./W
('a','h')
real 136m56.882s
user 135m13.690s
sys 0m2.370s
$
which is a speedup of around 7 or 8 percent.
> If you really need a reason which doesn't involve bottom, consider a
> (fairly common) call such as:
>
> zipWith f xs [1..]
>
> If xs is finite, your version of zipWith would evaluate the infinite
> list [1..] one place beyond that which was really needed.
Sure, there is a single extra amount of evaluation needed to work out if
there is a following list item (I guess this could be quite high in more
complex cases - is this the reason?), but there is a constant time
speedup on every zip(With) call that actually does some zipping.
Ian
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