[Haskell-cafe] how to break foldl' ?
Malcolm Wallace
Malcolm.Wallace at cs.york.ac.uk
Fri Sep 30 06:51:34 EDT 2005
> say if I want to sum a list of numbers but only until
> it hits a max limit.
>
> Currently, I control it through the function and
> basically do nothing when the max is hit. However, if
> the list is very long, would this mean the same
> function would be called for the rest of the list
> which can be a waste of cycle ? In an imperative
> language, I just break/return in the middle of the
> loop.
If you are using foldl or foldl', then yes, the definition tells you
that 'foldl' itself will be applied as many times as the length of
the list:
foldl f z [] = z
foldl f z (x:xs) = foldl f (f z x) xs
For your situation, foldr is better:
foldr f z [] = z
foldr f z (x:xs) = f x (foldr f z xs)
The function 'f' is the outermost application, therefore it can decide
to ignore its second argument, meaning that the recursive call to
foldr is never computed.
Regards,
Malcolm
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