using less stack
Sat, 23 Mar 2002 17:01:21 +0100
You don't have to define cpsfold explicitly recursively since it can be
expressed in terms of foldr:
cpsfold f a xs = foldr (\x k y -> f x y k) id xs a
The following definition would even be better (but not equivalent):
cpsfold' f a xs = foldr (\x k y -> f y x k) id xs a
The list members are now 'consumed' left-to-right by f, with initial value
answer4 = foldr (\x k a -> if x > a then k x else k a) id [1..500000] 0
also works without a crash or insufficient stack space.
----- Original Message -----
From: "Amanda Clare" <email@example.com>
To: "C.Reinke" <C.Reinke@ukc.ac.uk>
Sent: Wednesday, March 20, 2002 7:45 PM
Subject: Re: using less stack
> [cpsfold omitted]
> > Actually, and quite apart from it being cumbersome to use, I've got
> > my doubts about whether this cpsfold really does the job (is that
> > just me missing some point?-).
> It does the job for me! In practical terms I can see it works. I'm not
> an expert - I may have this all wrong, but perhaps the point you're
> looking for is that the arguments to f are brought to the very outside
> of the expression, and hence available for evaluation. Imagine for
> from foldr: f x1 (f x2 (f x3 (f x4 a)))
> from foldl: f (f (f (f a x1) x2) x3) x4
> Neither are available for immediate evaluation. In the case of foldr the
> end of the list (x4) has to be reached before anything can be evaluated.
> In the case of foldl the first function to be pulled upon is the
> outermost f, which then can't do anything useful (in my case) without
> evaluating its second argument, and so on.
> with the cpsfold I get
> f x1 a (\y1 -> f x2 y1 (\y2 -> f x3 y3 (\y3 -> f x4 y4 (\y4 -> y4)
> so x1 and a are available immediately for f to use, and f x1 a is the
> outermost expression so will be evaluated first.
> See for yourself with the following (difference can be seen in ghc with
> standard 1M stack):
> > answer1 = foldr larger 0 [1..500000]
> > answer2 = foldl larger 0 [1..500000]
> > answer3 = cpsfold cpslarger 0 [1..500000]
> > larger x y = if x > y then x else y
> > cpslarger x y k = if x > y then k x else k y
> > Also, I'm curious to know why the usual strict variant of foldl
> > doesn't help in this case?
> > foldl' f a  = a
> > foldl' f a (x:xs) = (foldl' f $! f a x) xs
> Because $! and seq only evaluates enough to make sure the answer is not
> bottom, and if my f is complex then it doesn't do enough.
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