Couldn't match expected type [a] with actual type `Integer -> [Integer]'

Question

I have the following code that finds the divisors of an Integer, then all the subsets of the divisors, then sums all the divisors in each subset, then tests to see if that Integer is represented in the list of summations. In findWeird this is performed over a list of Integers.

allSubsets :: (Integral a ) => [a] -> [[a]]
allSubsets [] = [[]]
allSubsets ( n : nr ) = allSubsets nr ++ map (n:) (allSubsets nr)

sumAllSubsets :: (Integral a ) => [a] -> [a]
sumAllSubsets s = map sum (allSubsets s)

allDivisors :: (Integral a) => a -> [a]
allDivisors 0 = []
allDivisors n   | n > 0 = [d | d <- [1..n], n `mod` d == 0]
                | n < 0 = -1 : allDivisors (-n)

findWeird :: (Integral a) => [a] -> [a]
findWeird [] = []
findWeird (n:nn) = (    if n `elem` (AS.sumAllSubsets (DFL.allDivisors))
                        then []
                        else [n]) ++ findWeird nn

Problem is that I get the error:

test.hs:15:61:

Couldn't match expected type `[a]'
            with actual type `Integer -> [Integer]'
In the first argument of `sumAllSubsets', namely `(allDivisors)'
In the second argument of `elem', namely
  `(sumAllSubsets (allDivisors))'
In the expression: n `elem` (sumAllSubsets (allDivisors))

But as far as know allDivisors produces a [Integral] and sumAllSubsets takes an [Integral], so I was just wondering if anyone could help. Thanks.

huon huon · Accepted Answer · 2014-06-30T12:15:14

I think the problem is you're not actually apply allDivisors to anything:

AS.sumAllSubsets (DFL.allDivisors)

is just applying sumAllSubsets to the function allDivisors, not to its [Integer] return value. Maybe you meant AS.sumAllSubsets (DFL.allDivisors n), that is, applying allDivisors to n?

(BTW, findWeird is just doing a filter, and can be written as

findWeird nn = filter (\n -> n `notElem` (sumAllSubsets $ allDivisors n)) nn

where I've also taken the liberty to reduce some nesting via the ($) operator.)

Couldn't match expected type [a] with actual type `Integer -> [Integer]'

1 Answers