[Haskell-cafe] Splitting a list

Steve Schafer steve at fenestra.com
Fri Apr 23 13:36:12 EDT 2004

On Wed, 21 Apr 2004 14:57:57 +0100, you wrote:

>How about implementing a directly recursive solution?  Simply
>accumulate the sum so far, together with the list elements you have
>already peeled off.  Once the sum plus the next element would exceed
>the threshold, emit the accumulated elements, and reset the sum
>to zero.
>    splitlist threshold xs = split 0 [] xs
>      where
>        split n acc [] = reverse acc: []
>        split n acc (x:xs)
>            | x >= threshold  = error (show x++" exceeds threshold ")
>            | n+x > threshold = reverse acc : split 0 [] (x:xs)
>            | otherwise       = split (n+x) (x:acc) xs

Thanks. Apart from a small off-by-one problem (the "x >= threshold" test
needs to be "x > threshold" instead), it works fine.

I had actually started along those lines, but got bogged down in the
details of passing the accumulator around, and ended up painting myself
into a corner, so I abandoned that approach (prematurely, as it turns

And thanks to everyone else who replied--I don't want to clutter the
list with a lot of individual replies. As you can probably tell, I've
only recently begun playing with Haskell, and the process of
reconfiguring my neurons into recursive loops has not yet been


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