On Mar 28, 2006, at 1:02 AM, Tomasz Zielonka wrote:
On Mon, Mar 27, 2006 at 03:10:18PM -0800, Greg Fitzgerald wrote:
hold a part of the data in memory while you show the first one,
Here would be a better example then.
f lst = show (sum (filter (> 1) lst), sum (filter (> 2) lst))
It ought to be *possible* to compute both operations without
holding onto
any of the list elements.
I wonder if it would be possible to remove the space-leak by
running both
branches concurrently, and scheduling threads in a way that would
minimise the space-leak. I proposed this before
http://www.haskell.org/pipermail/haskell-cafe/2005-December/
013428.html
I would like to hear opinions from some compiler gurus.
This is possible in principle with something like resource-bounded
eagerness, but it's not at all easy. The problem is this: when lst
gets big, you need to identify who's hanging on to it, and figure out
that they are actually planning to consume it and come up with
something smaller as a result. This is all pretty heavyweight---not
hard in principle, but hard enough in practice that it may not be
worth the investment.
That said, there's a transformation that goes something like this:
a = foldr f z xs ==> (a,b) = foldr (f `cross` g)
(z,y) xs
b = foldr g y xs
This could in principle at least pluck the lowest-hanging fruit (sum,
filter, etc.).
However it runs into some significant problems:
- Only works with folds
- Has some problems with bottoms, if I recall rightly
- Not expressible using something like RULES;
requires a special transformation in the compiler.
- It is often a pessimization.
That last point is a killer.
Best regards
Tomasz
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