Haskell: Parse Error in nested where clause

Question

I have the following code, which implements the sieve or Eratosthenes:

primeSieve :: Int -> [Int] -- Returns a list of primes up to upperbound
primeSieve upperbound = filter[2..upperbound] 
  where filter ls indx = 
    let divisor = (ls!!indx)
    let filtered = [x | x <- ls, x `mod` divisor /= 0]
    if divisor*divisor >= last filtered
      then filtered
      else filter filtered (indx+1)

I get a parse error on line 4, "possible incorrect indentation or mismatched brackets".

Why is this?

baxbaxwalanuksiwe · Accepted Answer · 2016-08-21 09:37:04Z

2

I believe you wanted to write filter with a do-notation. You could just add a do just after the filter ls indx =. However, this code being pure (i.e. non monadic), I would recommend this syntax:

primeSieve :: Int -> [Int] -- Returns a list of primes up to upperbound
primeSieve upperbound = filter [2..upperbound]
  where
    filter ls indx =
        let divisor = (ls!!indx)
            filtered = [x | x <- ls, x `mod` divisor /= 0]
        in if divisor*divisor >= last filtered
          then filtered
          else filter filtered (indx+1)

... but your code gives me the following error:

<file>:13:25:
Couldn't match expected type `[Int]'
            with actual type `Int -> [Int]'
Probable cause: `filter' is applied to too few arguments
In the expression: filter [2 .. upperbound]
In an equation for `primeSieve':
    primeSieve upperbound
      = filter [2 .. upperbound]
      where
          filter ls indx
            = let ...
              in
                if divisor * divisor >= last filtered then
                    filtered
                else
                    filter filtered (indx + 1)
Failed, modules loaded: none.

I think you meant to pass 0 as a second argument to filter:

primeSieve :: Int -> [Int] -- Returns a list of primes up to upperbound
primeSieve upperbound = filter [2..upperbound]
  where
    filter ls indx =
        let divisor = (ls!!indx)
            filtered = [x | x <- ls, x `mod` divisor /= 0]
        in if divisor*divisor >= last filtered
          then filtered
          else filter filtered (indx+1)

answered Aug 21, 2016 at 9:37

baxbaxwalanuksiwe

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2 Comments

Dr. John A Zoidberg Over a year ago

What is the meaning of the "in if"?

baxbaxwalanuksiwe Over a year ago

This is just a let ... in clause followed by a if expression. Nothing magic here.

mariop · Accepted Answer · 2016-08-21 09:51:21Z

1

The issue is the indentation of the definition of filter which is smaller than what it shoul be. The important part is

where filter ls indx = 
  let divisor = (ls!!indx)

where the let starts 4 characters before the filter on the line above. That triggers the syntax error. To fix it you can indent more the filter definition

where filter ls indx = 
        let divisor = (ls!!indx)

a more common format is

where
  filter ls indx = 
    let divisor = (ls!!indx)

because in this way you don't indent too much.

You can find more information about indentation on the haskell wiki which contains some nice visual examples about this kind of errors.

answered Aug 21, 2016 at 9:51

mariop

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