length of list using foldr in haskell

Question

I wrote the following code to calculate length of a list using foldr in haskell. When I compile the code it gives me error as "myfoldr.hs:3:1: parse error on input `where'". Can anyone please tell me what thing I might be missing or doing wrong in this code ?

mylength :: [Int] -> Int
mylength l = foldr f 0 l
where
f :: Int -> Int -> Int
f x y = y+1

Answer 1

In Haskell, whitespace matters - look at the guide on the Haskell wiki .

Formatting your code more correctly gives:

mylength :: [Int] -> Int
mylength l = foldr f 0 l
  where
    f :: Int -> Int -> Int
    f x y = y + 1

Which works perfectly (although the argument x to f is a bit redundant, you might want to write it as f _ y = y + 1 instead, or use a lambda expression like foldr (\\_ x -> x + 1) 0 l ).

Answer 2

This is an indentation error: you have to indent the where clause, since otherwise Haskell will see the definition of f as a separate function. So we can fix it with:

mylength :: [Int] -> Int
mylength l = foldr f 0 l
    where f :: Int -> Int -> Int
          f x y = y+1

Nevertheless we can still make it more generic: instead of defining it for an [Int] list, we can define it over an [a] list, with:

mylength :: [a] -> Int
mylength l = foldr f 0 l
    where f x y = y+1

We can also rewrite f as const (+1) , so:

mylength :: Num n => [a] -> n
mylength = foldr (const (1+)) 0

Note that we can apply an eta-reduction here: remove l both in the head and the body of the mylength definition. Or in case we know that the number is also enumerable , we can use succ instead of (1+) :

mylength :: (Enum n, Num n) => [a] -> n
mylength = foldr (const succ) 0