# foldl vs foldl'

Posted on December 21, 2016 by Kwang Yul Seo
Tags: fold, recursion

Chris Allen mentioned foldl as one of the newbie traps in Haskell.

foldl’ is always what you want, don’t use foldl!

Because foldl always has to examine the whole list, there is no reason to make it lazy. It just uses more memory to do the same thing as foldl'.

Due to the thunking behavior of foldl, it is wise to avoid this function in real programs: even if it doesn’t fail outright, it will be unnecessarily inefficient. Instead, import Data.List and use foldl’

Haskell Wiki compares foldr, foldl and foldl' and recommends using either foldr or foldl'.

foldl’ is the more efficient way to arrive at that result because it doesn’t build a huge thunk.

But here comes a question. If foldl' is almost always better than foldl, why do we have foldl anyway? It makes sense only when the combining function is non-strict in its first argument. (The example is taken from the Haskell Wiki.)

(?) :: Int -> Int -> Int
_ ? 0 = 0
x ? y = x*y

list :: [Int]
list = [2,3,undefined,5,0]

okey = foldl (?) 1 list
boom = foldl' (?) 1 list

Evaluation of okey:

okey -->
foldl (?) 1 [2,3,undefined,5,0] -->
foldl (?) (1 ? 2) [3,undefined,5,0] -->
foldl (?) ((1 ? 2) ? 3) [undefined,5,0] -->
foldl (?) (((1 ? 2) ? 3) ? undefined) [5,0] -->
foldl (?) ((((1 ? 2) ? 3) ? undefined) ? 5) [0] -->
foldl (?) (((((1 ? 2) ? 3) ? undefined) ? 5) ? 0) [] -->
((((1 ? 2) ? 3) ? undefined) ? 5) ? 0 -->
0

Evaluation of boom:

boom -->
foldl' (?) 1 [2,3,undefined,5,0] -->
1 ? 2 --> 2
foldl' (?) 2 [3,undefined,5,0] -->
2 ? 3 --> 6
foldl' (?) 6 [undefined,5,0] -->
6 ? undefined -->
*** Exception: Prelude.undefined

This example actually shows why foldl is so useless because it is hard to find a function which is non-strict in its first argument.

Many functions in Haskell are non-strict in its second argument and this is why foldr is useful. For example, (&&) is non-strict in its second argument and and can be efficiently defined using foldr.

(&&)                    :: Bool -> Bool -> Bool
True  && x              =  x
False && _              =  False

and                     :: [Bool] -> Bool
and                     =  foldr (&&) True

In conclusion, we should use foldl' by default unless we have a very compelling reason to use foldl instead.

But, wait! Let’s check how our beloved sum function is written. Because (+) is strict in both of its arguments, foldl' should have been used. But here’s the actual code taken from the base package.

sum                     :: (Num a) => [a] -> a
{-# INLINE sum #-}
sum                     =  foldl (+) 0

OMG! There is a historical accident here. Interested readers are referred to Fixing foldl article from Well-Typed.