haskell error: map is applied to too many arguments

Question

i keep getting this error and i cant work out why

sumSquares a = map (^2) a . foldr (+) 0

Im adding up the squares of a list of numbers.

it looks like you used (.) in wrong order. sumSquares = foldr (+) 0 . map (^2) — Sassa NF, Jan 15 '14 at 17:31

score 5 · Accepted Answer · answered Jan 15 '14 at 17:27

5

. is function composition, so what you're saying is, take the function foldr (+) 0 :: [Int] -> Int and compose it with map (^2) a :: [Int], but this makes no sense, your second "function" in this composition isn't really a function.

Would you'd want instead is to swap them and use application, not composition

 foldr (+) 0 (map (^2) a)
 foldr (+) 0 $ map (^2) a

and this can be converted back into composition with what's called "eta conversion"

 sumSquares = foldr (+) 0 . map (^2)
 sumSquares = sum . map (^2)

As an aside, foldr is probably the wrong fold here, foldl' would be more space efficient.

answered Jan 15 '14 at 17:27

daniel gratzer

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`sum` is also non-struct in its accumulator. – Tom Ellis Jan 16 '14 at 08:13
@TomEllis Yeah, which is why you can blow the stack with it, where as `foldl'` runs in constant space – daniel gratzer Jan 16 '14 at 14:36

Sibi · Answer 2 · 2014-01-15T17:38:32.740

3

You want to actually do this:

sumSquares a = foldr (+) 0 $ map (^2) a

Demo in ghci:

ghci> sumSquares [2,2]
8

To solve this problem using (.) operator, you can use the solution as mentioned by @Sassa:

sumSquares = foldr (+) 0 . map (^2)

But I find that hard to read, or you can see the jozefg solution where he uses sum and (.)

edited Jan 15 '14 at 17:38

answered Jan 15 '14 at 17:20

Sibi

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I have been told to use the . operator, why does this not work or how can I alter my code to use the dot? thanks – user3105607 Jan 15 '14 at 17:30
@user3105607 See jozefg answer. :) – Sibi Jan 15 '14 at 17:31

haskell error: map is applied to too many arguments

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