haskell parse error on input "="

Question

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

main = do
    a = [ 5, 1, 9, 4, 6, 7, 3]
    print quicksort a

I got this error:

[1 of 1] Compiling Main             ( quicksort.hs, quicksort.o )

quicksort.hs:10:11: parse error on input `='

Really don't undrestand why. Double checked this code several times. Still confusing. Need help. thanks!

daniel gratzer · Accepted Answer · 2014-06-25 09:05:05Z

3

You forgot a let,

quicksort :: (Ord a) => [a] -> [a]
quicksort [] = []
quicksort (x:xs) =
    let smallerOrEqual = [a | a <- xs, a <= x]
        larger = [a | a <- xs, a > x]
    in quicksort smallerOrEqual ++ [x] ++ quicksort larger -- ++ not +=

main = do
    let a = [ 5, 1, 9, 4, 6, 7, 3]
    print (quicksort a) -- The parens are needed too

I fixed a few other errors too.

It's worth noting that this quicksort isn't in place and therefore is not really quicksort, though it is pretty.

edited Jun 25, 2014 at 9:05

answered Jun 25, 2014 at 3:43

daniel gratzer

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8 Comments

ApproachingDarknessFish Over a year ago

Yeah, this quickSort implementation is pretty much Haskell's signature code snippet since it's so pretty.

ssm Over a year ago

Shouldnt it be in (quicksort smallerOrEqual) ++ [x] ++ (quicksort larger) ?

daniel gratzer Over a year ago

@ssm Function application has the highest precedence, it doesn't matter

BufBills Over a year ago

why is this not real quicksort? Not in place ? require more spaces,right? Other than that, this still implements quick sort algorithm IMO.

daniel gratzer Over a year ago

@ValekHalfHeart Nope, you just need to use the ST monad, this means you essentially are writing C with a nicer calling interface though :/

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ApproachingDarknessFish · Accepted Answer · 2014-06-25 03:48:19Z

3

Within do blocks, bindings must be preceded by the let keyword, so the line should be let a = [ 5, 1, 9, 4, 6, 7, 3].

There are a few other errors, however. You appear to have mistyped the ++ operator and you forgot to sort the larger section of the list. Your print statement won't work because print quicksort a is parsed as passing both the function quickSort and the value a to print. You can easily fix this using parentheses or the function application operator ($). This code works perfectly:

quicksort :: (Ord a) => [a] -> [a]
quicksort [] = []
quicksort (x:xs) =
    let smallerOrEqual = [a | a <- xs, a <= x]
        larger = [a | a <- xs, a > x]
    in quicksort smallerOrEqual ++ [x] ++ quickSort larger

main = do
    let a = [ 5, 1, 9, 4, 6, 7, 3]
    print $ quicksort a

edited Jun 25, 2014 at 3:48

answered Jun 25, 2014 at 3:42

ApproachingDarknessFish

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