Haskell quicksort: Parsing error with 'where'

I'm just starting to teach myself Haskell out of the book "Learn you a haskell for great good", and I rewote the quicksort in Chapter 5 using where:

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

but when I loaded it into GHCi 7.0.3, I got the f开发者_运维问答ollowing error:

parse error on input '='

The original code on the book:

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

Can you please help me find it why it doesn't work with where?

That's the whitespace rule. Your definitions after where have to be at the same whitespace indentation. This will compile:

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

You may want to read this.

As an addendum to cldy's answer, note that you can also format where-clauses like this:

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

I personally prefer this as it conserves some horizontal space, and because most editors can't intelligently auto-indent to the correct column when using the more traditional style.