Understanding Fold · David Raab

Understanding Fold

A very important function in List-processing is the List.fold function. List.fold is a very powerful function that gives you a lot of flexibility to do any kind of list transformation. The disadvantage is that this kind of power makes understanding List.fold a lot harder. In this article we look at some visualizations to better understand List.fold.

First we need to understand how we visualize a function. A function has some inputs and one output and we visualize it as a box with the inputs on the top and the output on the bottom. As an example this function and functions call:

let times2 x = x * 2

times2  5 // 10
times2  7 // 14
times2 10 // 20

Will be visualized like this:

The times2 functions as boxes

We visualze a list like [1;2;3;4;5] like that:

A list containing elements 1 to 5

We don't directly jump to List.fold, instead we work towards List.fold by first looking at List.map and then List.reduce.


We could visualize a call like List.map times2 [1;2;3;4;5] with the technique we have seen so far. Draw a box with two inputs and a list as output. But we are not interested in visualizing the List.map call itself, we want to visualize how List.map internally works.

We can think of List.map as a function that applies its first argument, a function, to every element of the list passed as the second argument.

Visualization for List.map example

List.map is a very useful function as executing a function for every element is often needed in programming. On the other hand it is the exact opposite of powerful, it is a very specific function that only can manipulate one element of a list at a time. For example we cannot add all values of a list together, filter a list and do other kind of things with a list.


To understand List.reduce I use an example that is often used by others. First let me say that I don't think this is the best way to explain List.reduce but it is still a good start.

List.reduce (+) [1;2;3;4;5] // 15

The result of the above statement is 15. You can imagine that List.reduce just puts the + between every element in the list:

1 + 2 + 3 + 4 + 5

It isn't the best explanation because as it doesn't explain what this means:

List.reduce someFunc [1;2;3;4;5]

Putting someFunc between every element doesn't seem to make any sense!

// What does that mean?
1 someFunc 2 someFunc 3 someFunc 4 someFunc 5

But we ignore this for a moment.

First, we need to understand that + is also just a function. For example we could visualize 7 + 9 like this:

Visualization of: 7 + 9

Second, if we calculate 1 + 2 + 3 + 4 + 5 in our head then we don't do it in one big step. We actually do a lot of steps and create a lot of intermediate results.

We first calculate 1 + 2 and keep 3 in our head. Then we calculate 3 + 3 and keep 6 in our head. Then we calculate 6 + 4 and finally we do 10 + 5 in our head. If we add parenthesis around every step we do something like this in our head:

((((1 + 2) + 3) + 4) + 5)

This idea of doing multiple calculations and remembering an intermediate value after each step is exactly how List.reduce works.

Visualization of reduce

This visualization also explains how List.reduce works if we pass it other functions instead of operators like +.

List.reduce max [4;12;18;7;3] // 18
Visualization of reduce with max

We also can write it with parenthesis:

(max (max (max (max 4 12) 18) 7) 3)

This is how List.reduce works, but also List.reduce has some limitations.

First, what should List.reduce do if you only pass a list with one or zero elements? In the case of one argument it just returns this one element without that functions like + or max is executed once. In the case of zero arguments it throws an exception.

A second limitation is that we are still limited in which kind of list transformations we can create. This becomes more obvious if we look at the types of the function that we must pass to List.reduce. We must pass a function where all input and output types are the same and must be of the type that the list contains.

This becomes obvious if we look at how List.reduce (+) [1;2;3;4;5] works. The first calculation it does is 1 + 2. So it passes 1 as the first argument and 2 as the second argument to +. As both values come from the same list and all values in a list must be of the same type it means both input values must be the exact same type of the list.

As the result of 1 + 2 is again passed as the first argument to the next + call it also means the output of a function must be the same as its input type.

This means when we pass a list of int to List.reduce then we must pass a int -> int -> int function to List.reduce and overall it also returns an int as the final value. More general, we need a function of type:

a -> a -> a

Another way to describe how List.reduce works is to say it combines the first two values of a list into a new single value and repeats doing this step until we end up with a single value that is then returned.

Another visualization of reduce

The first visualization we have seen is more accurate, but also this kind of idea leads you to the correct result of any List.reduce call.


Finally we are at List.fold. The way how List.fold works is pretty much the exact same as List.reduce! The only difference is that we pass an initial starting value. Doesn't seems like a big deal but this initial value fixes all limitations of List.reduce.

List.fold (+) 0 [1;2;3;4;5] // 15

The List.fold call above internally does something like this:

Visualization of fold

Overall List.fold does one additional calculation compared to List.reduce because we pass it an initial value. But because of this it also always can return a value even if we pass it an empty list. This way it doesn't need to throw an exception.

The biggest advantage is that we can use different types. As the initial value is not part of the list it also can be another type.

The initial value is passed as the first argument to the folder function, the folder function then returns a new intermediate result that is used for the next call. This tells us that the first argument and the return value must be of the same type.

Only the second argument of the folder function must be the type thats inside of our list.

List.fold annotated

Let's say we want to traverse a list and build a string where every number is just concatenated. Then we already know that the initial value must be a string. The initial value must be a starting value, so we use the empty string. Next we know that the the first argument of the folder function is the initial value or any intermediate string, the second argument is one element from the list, and we must return the next intermediate string (or final return value).

List.fold (fun acc x -> acc + (string x)) "" [1;2;3;4;5] // "12345"

The execution of the above function looks like this:

Building a string

Instead of initial or intermediate value we use the term accumulator, hence the name acc in the anonymous function.

Now i want you to look at the following code:

let mutable acc = ""
for x in [1;2;3;4;5] do
    acc <- acc + (string x)

Can you see the similarities with the List.fold function call?

Comparing fold with a loop

In some way we can say that List.fold for immutable data-types is what a for-loop is for mutable data-types. List.fold is what you get if you try to eliminate all mutable variables and data-types. The whole purpose of looping is to mutate some state that is usually defined outside of the loop. In List.fold on the other hand the new state is always passed as an argument and we need to return the new state for the next function call.

It becomes more obvious if we go through the states of acc in the loop based code. We start with the empty string "". In the first loop iteration x is assigned 1 and we append this to acc and acc is set to "1". The second loop iteration assigns "12" to acc and so on. These are exactly the states our folder function return in the List.fold example.

You also can easily convert other loop code to a List.fold. As an example:

let mutable amountOfEvenNumbers = 0
let mutable sumOfEvenNumbers    = 0
for x in 1 .. 10 do
    if x % 2 = 0 then
        amountOfEvenNumbers <- amountOfEvenNumbers + 1
        sumOfEvenNumbers    <- sumOfEvenNumbers + x

amountOfEvenNumbers // 5
sumOfEvenNumbers    // 30 

Written as a fold:

let folder (amount,sum) x =
    if   x % 2 = 0
    then (amount+1, sum+x)
    else (amount, sum)

let (amount, sum) = List.fold folder (0,0) [1..10]

amount // 5
sum    // 30

It's a bit different. As a List.fold only supports one initial value we use a Tuple with two values. In the mutation based code we don't need an else branch as nothing happens with an odd number. In the List.fold example we must explicitly return the unchanged state.


As List.fold is basically just a for-loop it should now become clear why it is so powerful, yet some task are harder as it seems. As an example we want to build our own List.map with the help of List.fold.

As the return value is a list our initial value also must be a list and we start with the empty list as the initial value. Then we execute a user-defined function for every element and the result must be added to the empty list.

We could write something like that:

let map' f list =
    let folder acc x =
        let newElement = f x
        newElement :: acc
    List.fold folder [] list

map' times2 [1..5] // [10; 8; 6; 4; 2]

The problem is how we build the list. The cons operator :: only can prepend an element to a list, not append. Because of this we get [10;8;6;4;2] instead of the expected [2;4;6;8;10].

It is possible to write an append function that appends a single element to a list. The general problem is that this is a very inefficient operation with an immutable list. Whenever you append an element a whole new list must be created and all elements must be copied.

A better approach would be to traverse the list in reverse order. We start at the end (the right side) and prepend the result to an empty list. This is excactly how List.foldBack works.

let map'' f list =
    let folder x acc =
        let newElement = f x
        newElement :: acc
    List.foldBack folder list []

map'' times2 [1..5] // [2;4;6;8;10]
FoldBack that shows the map implementation

If you look at the visualization it also becomes clear why the order of the initial value is different. In List.fold you start from the left and the initial value is also places on the left-side from the list.

In List.foldBack we start at the right and we also place the initial value at the right-side of the list. The List.fold and List.foldBack resembles that idea.

List.fold     folder initialValue list
List.foldBack folder list         initialValue

Also the folder function receives the accumulator as the second argument (right argument) in List.foldback while List.fold receives it as the first argument (left argument).


The List.fold and List.foldBack functions are powerful function, but this doesn't mean they are "good". In general you should avoid powerful language constructs. Functions like map or filter are overall less powerful but many times easier to understand. This is also the idea of Structured Programming.

But avoiding doesn't mean: "Never use it". If you need a special list manipulation and there doesn't exists a built-in function then use List.fold or List.foldBack and create the needed function on your own.

Besides list manipulation (list as input and list as output), List.fold and List.foldBack are the best functions if you want to convert a list to any other non-list data-types.

If you are interested in more code examples and you want to know how to implement List.fold and List.foldBack yourself you also can read From mutable loops to immutable folds.

Nowadays fold and foldBack functions are not only implemented for the List data-type. The idea was generalized and is known as Catamorphisms. You should know about Algebraic Data-Types especially recursive data-types beforehand to understand them.

module Main
val times2 : x:int -> int

Full name: Main.times2
val x : int
Multiple items
module List

from Microsoft.FSharp.Collections

type List<'T> =
  | ( [] )
  | ( :: ) of Head: 'T * Tail: 'T list
  interface IEnumerable
  interface IEnumerable<'T>
  member GetSlice : startIndex:int option * endIndex:int option -> 'T list
  member Head : 'T
  member IsEmpty : bool
  member Item : index:int -> 'T with get
  member Length : int
  member Tail : 'T list
  static member Cons : head:'T * tail:'T list -> 'T list
  static member Empty : 'T list

Full name: Microsoft.FSharp.Collections.List<_>
val reduce : reduction:('T -> 'T -> 'T) -> list:'T list -> 'T

Full name: Microsoft.FSharp.Collections.List.reduce
val max : e1:'T -> e2:'T -> 'T (requires comparison)

Full name: Microsoft.FSharp.Core.Operators.max
val fold : folder:('State -> 'T -> 'State) -> state:'State -> list:'T list -> 'State

Full name: Microsoft.FSharp.Collections.List.fold
val acc : string
Multiple items
val string : value:'T -> string

Full name: Microsoft.FSharp.Core.Operators.string

type string = System.String

Full name: Microsoft.FSharp.Core.string
val mutable acc : string

Full name: Main.acc
val mutable amountOfEvenNumbers : int

Full name: Main.amountOfEvenNumbers
val mutable sumOfEvenNumbers : int

Full name: Main.sumOfEvenNumbers
val x : int32
val folder : amount:int * sum:int -> x:int -> int * int

Full name: Main.folder
val amount : int
val sum : int
val amount : int

Full name: Main.amount
val sum : int

Full name: Main.sum
val map' : f:('a -> 'b) -> list:'a list -> 'b list

Full name: Main.map'
val f : ('a -> 'b)
Multiple items
val list : 'a list

type 'T list = List<'T>

Full name: Microsoft.FSharp.Collections.list<_>
val folder : ('b list -> 'a -> 'b list)
val acc : 'b list
val x : 'a
val newElement : 'b
val map'' : f:('a -> 'b) -> list:'a list -> 'b list

Full name: Main.map''
val folder : ('a -> 'b list -> 'b list)
val foldBack : folder:('T -> 'State -> 'State) -> list:'T list -> state:'State -> 'State

Full name: Microsoft.FSharp.Collections.List.foldBack
type 'T list = List<'T>

Full name: Microsoft.FSharp.Collections.list<_>