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Home/ Questions/Q 1009175
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T08:53:04+00:00 2026-05-16T08:53:04+00:00

Does anybody know the steps of haskell ‘foldr’ use of function? GHCI Command Window:

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Does anybody know the steps of haskell ‘foldr’ use of function?

GHCI Command Window:

foldr (\x y -> 2*x + y) 4 [5,6,7] 

The result after evaluation:

40

Steps on this,

Prelude> foldr (\x y -> 2*x + y) 4 [5,6,7] 
6 * 2 + (7 * 2 + 4)
12 + 18 = 30
5 * 2 + 30 = 40 v
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  1. Editorial Team
    Editorial Team
    2026-05-16T08:53:05+00:00Added an answer on May 16, 2026 at 8:53 am

    One definition of foldr is:

    foldr            :: (a -> b -> b) -> b -> [a] -> b
    foldr f acc []     = acc
    foldr f acc (x:xs) = f x (foldr f acc xs)
    

    The wikibook on Haskell has a nice graph on foldr (and on other folds, too):

      :                         f
     / \                       / \
    a   :       foldr f acc   a   f
       / \    ------------->     / \
      b   :                     b   f
         / \                       / \
        c  []                     c   acc
    

    I.e. a : b : c : [] (which is just [a, b, c]) becomes f a (f b (f c acc)) (again, taken from wikibook).

    So your example is evaluated as let f = (\x y -> 2*x + y) in f 5 (f 6 (f 7 4)) (let-binding only for brevity).

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