I tried to make a program in Haskell that returns the sum of squares of only the positive numbers in a list by using the filter function.
this is my first attempt:
sumsq :: [Int] -> Int
sumsq xs = foldr (+) 0 filter isPositive xs
isPositive :: Int -> Bool
isPositive x | x > 0 = True
| otherwise = False
but it's not working and this is the error message I am getting :
Couldn't match expected type `[(Int -> Bool) -> [Int] -> Int]'
with actual type `(a0 -> Bool) -> [a0] -> [a0]'
In the third argument of `foldr', namely `filter'
In the expression: foldr (+) 0 filter isPositive xs
In an equation for ` sumsq':
sumsq xs = foldr (+) 0 filter isPositive xs
solution: after adding parentheses it's working correctly.
sumsq :: [Int] -> Int
sumsq xs = foldr (+) 0 (filter isPositive xs)
isPositive :: Int -> Bool
isPositive x | x > 0 = True
my new question is: why is it working with the parentheses and not without the parentheses ?
Without the parentheses, you're saying that filter, isPositive, and xs are all arguments to the fold. That doesn't work because foldr expects a list as its third argument, not two functions followed by a list. What you want is to pass the result of calling filter isPositive xs as the third argument to foldr, and that's what the parentheses specify.
You can get the same effect by writing foldr (+) 0 $ filter isPositive xs, since the $ operator basically means to wrap a set of parentheses around everything that follows it.
Note that you can also write this function more concisely in point-free form (no need to mention xs) and without an explicit fold, using function composition:
sumsq = sum . map (^2) . filter (>0)
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