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Home/ Questions/Q 8890555
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T22:31:45+00:00 2026-06-14T22:31:45+00:00

In order to improve my Haskell skills I decided to go through some example

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In order to improve my Haskell skills I decided to go through some example code and try to rewrite in a different way.

Here is the initial function:

quicksort :: (Ord a) => [a] -> [a]  
quicksort [] = []  
quicksort (x:xs) =   
    let smallerSorted = quicksort [a | a <- xs, a <= x]  
        biggerSorted = quicksort [a | a <- xs, a > x]  
    in  smallerSorted ++ [x] ++ biggerSorted  

Here is the migrated function:

quicksort :: (Ord a) => [a] -> [a]  
quicksort [] = []  
quicksort (x:xs) = smallerSorted ++ [x] ++ biggerSorted 
    where smallerSorted = quicksort [a | a <- xs, a <= x]  
        biggerSorted = quicksort [a | a <- xs, a > x]  

However looks to me that it is not working.
Is there something wrong? Can functions using LET ported to WHERE?

This is my error output:

/Users/graph/Documents/Uni/Haskell/hey.hs:43:5:
    parse error on input `biggerSorted'
Failed, modules loaded: none.

Many thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-14T22:31:47+00:00Added an answer on June 14, 2026 at 10:31 pm

    You need to match the indentation of the expressions in a where clause, e.g.

    quicksort :: (Ord a) => [a] -> [a]  
    quicksort [] = []  
    quicksort (x:xs) = smallerSorted ++ [x] ++ biggerSorted 
        where smallerSorted = quicksort [a | a <- xs, a <= x]  
              biggerSorted = quicksort [a | a <- xs, a > x]  
    

    or

    quicksort :: (Ord a) => [a] -> [a]  
    quicksort [] = []  
    quicksort (x:xs) = smallerSorted ++ [x] ++ biggerSorted 
        where
            smallerSorted = quicksort [a | a <- xs, a <= x]  
            biggerSorted = quicksort [a | a <- xs, a > x]  
    
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