# Foldr Foldl Foldl'

### From HaskellWiki

To *foldr*, *foldl* or *foldl'* that's the question! This article demonstrates the differences between these different folds by a simple example.

If you want you can copy/paste this article into your favorite editor and run it.

We are going to define our own folds so we hide the ones from the Prelude:

> import Prelude hiding (foldr, foldl)

## 1 Foldr

Say we want to calculate the sum of a very big list:

> veryBigList = [1..1000000]

Lets start with the following:

> foldr f z [] = z > foldr f z (x:xs) = x `f` foldr f z xs > sum1 = foldr (+) 0 > try1 = sum1 veryBigList

If we evaluate *try1* we get:

`*** Exception: stack overflow`

Too bad... So what happened:

try1 --> sum1 veryBigList --> foldr (+) 0 veryBigList --> foldr (+) 0 [1..1000000] --> 1 + (foldr (+) 0 [2..1000000]) --> 1 + (2 + (foldr (+) 0 [3..1000000])) --> 1 + (2 + (3 + (foldr (+) 0 [4..1000000]))) --> 1 + (2 + (3 + (4 + (foldr (+) 0 [5..1000000])))) --> -- ... -- ... My stack overflows when there's a chain of around 500000 (+)'s !!! -- ... But the following would happen if you got a large enough stack: -- ... 1 + (2 + (3 + (4 + (... + (999999 + (foldr (+) 0 [1000000]))...)))) --> 1 + (2 + (3 + (4 + (... + (999999 + (1000000 + ((foldr (+) 0 []))))...)))) --> 1 + (2 + (3 + (4 + (... + (999999 + (1000000 + 0))...)))) --> 1 + (2 + (3 + (4 + (... + (999999 + 1000000)...)))) --> 1 + (2 + (3 + (4 + (... + 1999999 ...)))) --> 1 + (2 + (3 + (4 + 500000499990))) --> 1 + (2 + (3 + 500000499994)) --> 1 + (2 + 500000499997) --> 1 + 500000499999 --> 500000500000

The problem is that (+) is strict in both of its arguments. This means that both arguments must be fully evaluated before (+) can return a result. So to evaluate:

1 + (2 + (3 + (4 + (...))))

`1` is pushed on the stack. Then:

2 + (3 + (4 + (...)))

is evaluated. So `2` is pushed on the stack. Then:

3 + (4 + (...))

is evaluated. So `3` is pushed on the stack. Then:

4 + (...)

is evaluated. So `4` is pushed on the stack. Then: ...

... your limited stack will eventually run full when you evaluate a large enough chain of (+)s. This then triggers the stack overflow exception.

Lets think about how to solve it...

## 2 Foldl

One problem with the chain of (+)'s is that it can't be made smaller (reduced) until the very last moment, when it's already too late.

The reason we can't reduce it is that the chain doesn't contain an
expression which can be reduced (a *redex*, for **red**ucible
**ex**pression.) If it did we could reduce that expression before going
to the next element.

We can introduce a redex by forming the chain in another way. If
instead of the chain `1 + (2 + (3 + (...)))` we could form the chain
`(((0 + 1) + 2) + 3) + ...`, then there would always be a redex.

We can form such a chain by using a function called *foldl*:

> foldl f z [] = z > foldl f z (x:xs) = let z' = z `f` x > in foldl f z' xs > sum2 = foldl (+) 0 > try2 = sum2 veryBigList

Lets evaluate *try2*:

`*** Exception: stack overflow`

Good Lord! Again a stack overflow! Lets see what happens:

try2 --> sum2 veryBigList --> foldl (+) 0 veryBigList --> foldl (+) 0 [1..1000000] --> let z1 = 0 + 1 in foldl (+) z1 [2..1000000] --> let z1 = 0 + 1 z2 = z1 + 2 in foldl (+) z2 [3..1000000] --> let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 in foldl (+) z3 [4..1000000] --> let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 z4 = z3 + 4 in foldl (+) z4 [5..1000000] --> -- ... after many foldl steps ... let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 z4 = z3 + 4 ... z999997 = z999996 + 999997 in foldl (+) z999997 [999998..1000000] --> let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 z4 = z3 + 4 ... z999997 = z999996 + 999997 z999998 = z999997 + 999998 in foldl (+) z999998 [999999..1000000] --> let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 z4 = z3 + 4 ... z999997 = z999996 + 999997 z999998 = z999997 + 999998 z999999 = z999998 + 999999 in foldl (+) z999999 [1000000] --> let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 z4 = z3 + 4 ... z999997 = z999996 + 999997 z999998 = z999997 + 999998 z999999 = z999998 + 999999 z100000 = z999999 + 1000000 in foldl (+) z1000000 [] --> let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 z4 = z3 + 4 ... z999997 = z999996 + 999997 z999998 = z999997 + 999998 z999999 = z999998 + 999999 z100000 = z999999 + 1000000 in z1000000 --> -- Now a large chain of +'s will be created: let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 z4 = z3 + 4 ... z999997 = z999996 + 999997 z999998 = z999997 + 999998 z999999 = z999998 + 999999 in z999999 + 1000000 --> let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 z4 = z3 + 4 ... z999997 = z999996 + 999997 z999998 = z999997 + 999998 in (z999998 + 999999) + 1000000 --> let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 z4 = z3 + 4 ... z999997 = z999996 + 999997 in ((z999997 + 999998) + 999999) + 1000000 --> let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 z4 = z3 + 4 ... in (((z999996 + 999997) + 999998) + 999999) + 1000000 --> -- ... -- ... My stack overflows when there's a chain of around 500000 (+)'s !!! -- ... But the following would happen if you got a large enough stack: -- ... let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 z4 = z3 + 4 in (((((z4 + 5) + ...) + 999997) + 999998) + 999999) + 1000000 --> let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 in ((((((z3 + 4) + 5) + ...) + 999997) + 999998) + 999999) + 1000000 --> let z1 = 0 + 1 z2 = z1 + 2 in (((((((z2 + 3) + 4) + 5) + ...) + 999997) + 999998) + 999999) + 1000000 --> let z1 = 0 + 1 in ((((((((z1 + 2) + 3) + 4) + 5) + ...) + 999997) + 999998) + 999999) + 1000000 --> (((((((((0 + 1) + 2) + 3) + 4) + 5) + ...) + 999997) + 999998) + 999999) + 1000000 --> -- Now we can actually start reducing: ((((((((1 + 2) + 3) + 4) + 5) + ...) + 999997) + 999998) + 999999) + 1000000 --> (((((((3 + 3) + 4) + 5) + ...) + 999997) + 999998) + 999999) + 1000000 --> ((((((6 + 4) + 5) + ...) + 999997) + 999998) + 999999) + 1000000 --> (((((10 + 5) + ...) + 999997) + 999998) + 999999) + 1000000 --> ((((15 + ...) + 999997) + 999998) + 999999) + 1000000 --> (((499996500006 + 999997) + 999998) + 999999) + 1000000 --> ((499997500003 + 999998) + 999999) + 1000000 --> (499998500001 + 999999) + 1000000 --> 499999500000 + 1000000 --> 500000500000 -->

Well, you clearly see that the redexes are created. But instead of being directly reduced, they are allocated on the heap:

let z1 = 0 + 1 z2 = z1 + 2 z3 = z2 + 3 z4 = z3 + 4 ... z999997 = z999996 + 999997 z999998 = z999997 + 999998 z999999 = z999998 + 999999 z1000000 = z999999 + 1000000 in z1000000

Note that your heap is only limited by the amount of memory in your system (RAM and swap). So the only thing this does is filling up a large part of your memory.

The problem starts when we finally evaluate z1000000:

We must evaluate `z1000000 = z999999 + 1000000`, so `1000000` is pushed on the stack. Then `z999999` is evaluated; `z999999 = z999998 + 999999`, so `999999` is pushed on the stack. Then `z999998` is evaluated; `z999998 = z999997 + 999998`, so `999998` is pushed on the stack. Then `z999997` is evaluated...

...your stack will eventually fill when you evaluate a large enough chain of (+)'s. This then triggers the stack overflow exception.

But this is exactly the problem we had in the foldr case — only now the chain of (+)'s is going to the left instead of the right.

So why doesn't the chain reduce sooner than before?

It's because of GHC's lazy reduction strategy: expressions are reduced only when they are actually needed. In this case, the outer-left-most redexes are reduced first. In this case it's the outer `foldl (+) ... [1..10000]`
redexes which are repeatedly reduced. So the inner `z1, z2, z3, ...` redexes only get reduced when the foldl is completely gone.

## 3 Foldl'

We somehow have to tell the system that the inner redex should be
reduced before the outer. Fortunately this is possible with the
*seq* function:

seq :: a -> b -> b

*seq* is a primitive system function that when applied to *x* and
*y* will first reduce *x* then return *y*. The idea is that *y* references *x* so that when *y* is reduced *x* will not be a big unreduced chain anymore.

Now lets fill in the pieces:

> foldl' f z [] = z > foldl' f z (x:xs) = let z' = z `f` x > in seq z' $ foldl' f z' xs > sum3 = foldl' (+) 0 > try3 = sum3 veryBigList

If we now evaluate *try3* we get the correct answer and we get it very quickly:

`500000500000`

Lets see what happens:

try3 --> sum3 veryBigList --> foldl' (+) 0 veryBigList --> foldl' (+) 0 [1..1000000] --> foldl' (+) 1 [2..1000000] --> foldl' (+) 3 [3..1000000] --> foldl' (+) 6 [4..1000000] --> foldl' (+) 10 [5..1000000] --> -- ... -- ... You see that the stack doesn't overflow -- ... foldl' (+) 499999500000 [1000000] --> foldl' (+) 500000500000 [] --> 500000500000

You can clearly see that the inner redex is repeatedly reduced first.

## 4 Conclusion

Usually the choice is between*first*argument,

> (?) :: 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

Let's see what happens:

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 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

*top-most constructor*. If the accumulator is a more complex object, then

*Real World Haskell*chapter 25.

## 5 Rules of Thumb for Folds

Folds are among the most useful and common functions in Haskell. They are an often-superior replacement for what in other language would be loops, but can do much more. Here are a few rules of thumb on which folds to use when.

*right*fold to use, in particular when transforming lists (or other foldables) into lists with related elements in the same order. Notably,

- conceptually reverses the order of the list. One consequence is that afoldl'(unlikefoldl') applied to an infinite list will be bottom; it will not produce any usable results, just as an expressfoldrwould not. Note thatreverse.foldl' (flip (:)) []==reverse
- often has much better time and space performance than afoldl'would for the reasons explained in the previous sections.foldr

- When the list to which it is applied is large, but definitely finite, you do not care about the implicit reversal (for example, because your combining function is commutative like ,(+), or(*)), and you seek to improve the performance of your code.Set.union
- When you actually do want to reverse the order of the list, in addition to possibly performing some other transformation to the elements. In particular, if you find that you precede or follow your fold with a reverse, it is quite likely that you could improve your code by using the other fold and taking advantage of the implicit reverse.

*short-circuit*, that is, terminate early by yielding a result which does not depend on the value of the accumulating parameter. When such possibilities arise with some frequency in your problem, short-circuiting can greatly improve your program's performance. Left folds can never short-circuit.