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Asked: May 11, 20262026-05-11T01:11:51+00:00 2026-05-11T01:11:51+00:00

Firstly, Real World Haskell , which I am reading, says to never use foldl

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Firstly, Real World Haskell, which I am reading, says to never use foldl and instead use foldl'. So I trust it.

But I’m hazy on when to use foldr vs. foldl'. Though I can see the structure of how they work differently laid out in front of me, I’m too stupid to understand when ‘which is better.’ I guess it seems to me like it shouldn’t really matter which is used, as they both produce the same answer (don’t they?). In fact, my previous experience with this construct is from Ruby’s inject and Clojure’s reduce, which don’t seem to have ‘left’ and ‘right’ versions. (Side question: which version do they use?)

Any insight that can help a smarts-challenged sort like me would be much appreciated!

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  1. 2026-05-11T01:11:52+00:00Added an answer on May 11, 2026 at 1:11 am

    The recursion for foldr f x ys where ys = [y1,y2,...,yk] looks like

    f y1 (f y2 (... (f yk x) ...)) 

    whereas the recursion for foldl f x ys looks like

    f (... (f (f x y1) y2) ...) yk 

    An important difference here is that if the result of f x y can be computed using only the value of x, then foldr doesn’t’ need to examine the entire list. For example

    foldr (&&) False (repeat False) 

    returns False whereas

    foldl (&&) False (repeat False) 

    never terminates. (Note: repeat False creates an infinite list where every element is False.)

    On the other hand, foldl' is tail recursive and strict. If you know that you’ll have to traverse the whole list no matter what (e.g., summing the numbers in a list), then foldl' is more space- (and probably time-) efficient than foldr.

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