How do I write a constant-space length function in Haskell?

Question

The canonical implementation of length :: [a] -> Int is:

length [] = 0
length (x:xs) = 1 + length xs

which is very beautiful but suffers from stack overflow as it uses linear space.

The tail-recursive version:

length xs = length' xs 0
  where length' [] n = n
        length' (x:xs) n = length xs (n + 1)

doesn't suffer from this problem, but I don't understand how this can run in constant space in a lazy language.

Isn't the runtime accumulating numerous (n + 1) thunks as it moves through the list? Shouldn't this function Haskell to consume O(n) space and lead to stack overflow?

(if it matters, I'm using GHC)

Answer 1

Yes, you've run into a common pitfall with accumulating parameters. The usual cure is to force strict evaluation on the accumulating parameter; for this purpose I like the strict application operator $!. If you don't force strictness, GHC's optimizer might decide it's OK for this function to be strict, but it might not. Definitely it's not a thing to rely on—sometimes you want an accumulating parameter to be evaluated lazily and O(N) space is just fine, thank you.

How do I write a constant-space length function in Haskell?

As noted above, use the strict application operator to force evaluation of the accumulating parameter:

clength xs = length' xs 0
  where length' []     n = n
        length' (x:xs) n = length' xs $! (n + 1)

The type of $! is (a -> b) -> a -> b, and it forces the evaluation of the a before applying the function.