# Haskell Datatypes and Folds: Part I

## Dec 17, 2009 00:00 · 1576 words · 8 minutes read haskell fold

Welcome to this little explanation on how to determine the fold of a Haskell datatype. First we’ll look at how we define functions over lists, something everyone starting with Haskell should be sufficiently familiar with, after which we move on to the datatypes. You’ll see different ways how to calculate the sum of a list, how to fold over a list, what datatypes are and, how to fold over a datatype, specifically the `BinTree a` datatype. Most importantly, I hope you will grow to understand what a fold is, and why they are so important and useful when programming Haskell.

During our functional programming course at Utrecht University I noticed students having anxiety of datatypes and even more for folds. Not because they couldn’t grasp the workings of a single example but more a lack of a view on the complete picture and lack of experience with the Haskell datatype way.

So, what are datatypes? Datatypes are a way of notating the abstract structure of your data. There are several datastructures known to man, such as lists and trees.

### Lists

Lists in Haskell are used in several ways. Today we will look at how to calculate the sum of a list. Intuitively you calculate the sum of a list by adding its elements together. Starting with the first element and then continuing on to the rest. This is how we literally translate that thought into Haskell code:

``````sum :: [Int] -> Int
sum []     = 0
sum (x:xs) = x + sum xs
-- sum [1..4] = 10``````

Would we be calculating the product of the list, we’d do the same except we multiply instead of adding.

``````product :: [Int] -> Int
product []     = 1
product (x:xs) = x * product xs
-- product [1..4] = 24``````

Looking at these two examples we can see that we have two similarities. We always have recursion on the tail of the list and we do something with the head of the list and the result of the recursion. In the first example we add them together, and in product we multiply them. Another property of most functions over lists is that there is a base case for the empty list. We call this the identity of our function, sometimes also referred to as unit. The identity of addition is 0 and the identity of multiplication is 1.

Now we look at one of the folds over lists defined in the prelude, `foldr`. Make sure you know what each parameter stands for. Note: There are some downsides to this function, mostly that it will not work for large lists, you’ll see more on this later.

``````foldr :: (a -> b -> b) -> b -> [a] -> b
foldr f z []     = z
foldr f z (x:xs) = f x (foldr f z xs)
-- The line above can also be written as:
-- foldr f z (x:xs) = x `f` (foldr f z xs)``````

Now, take a moment to let this function soak in and try to think of how you would define `sum` and `product` in terms of this `foldr`. Crucial at this point is to notice that we do not see any type hardcoded in the type of `foldr` It may be helpful to look at this fold, which is the identity function for lists:

``````idList :: [a] -> [a]
idList list = foldr (:) [] list
-- Recall: [1,2,3,4,5] == 1 : 2 : 3 : 4 : 5 : []``````

If we look at our sum function, the operator between each recursive call is `(+)` and our base case is 0. So that’s what we are going to use for our sum in terms of foldr.

``````sumList :: [Int] -> Int
sumList list = foldr (+) 0 list
-- sumList [1..4] = 10``````

Thus far we only have lists of `Int` for our examples. However, for sake of usability we will now move on to lists of numeric elements. Because `(+)` is defined for all numbers. (If you wish to read more on this subject please look up classes and instances.) Notice how the type of sumList changes while it’s definition remains unaltered.

``````sumList :: Num a => [a] -> a
sumList list = foldr (+) 0 list``````

Now that we have seen how we can determine the function and identity for our fold from our recursive function to a definition in terms of `foldr`. And most importantly, the foldr takes care of the recursive nature of the list for us and the only thing it asks us in return for that is an operator and an identity for your operation. So the fold can now be used to define several operations on lists.

I told you about a problem of `foldr`. Depending on the size of your list the above `sumList` may not work. Try the following on your machine:

``````\$ ghci
Prelude> let sumList = foldr (+) 0
Prelude> sumList [1..1000000]
*** Exception: stack overflow``````

Now try:

``````\$ ghci
Prelude> let sumList = Data.List.foldl' (+) 0
Prelude> sumList [1..1000000]
500000500000``````

For a complete overview and analysis on `foldr` and `foldl` please read A tutorial on the universality and expressiveness of fold, by Graham Hutton.

### Trees

First, let’s take a look at how data structures in Haskell can be defined. In short we have the `data` keyword, followed by zero or more ```type variables```, a `=` and then a number of ```data constructors``` separated by a `|`.

Now that you have familiarized yourself with lists we can proceed to a slightly more complicated datastructure. The Tree. In this example we will use a simple binary tree. A binary tree can be denoted as follows:

``data BinTree a = Node a (BinTree a) (BinTree a) | Leaf deriving (Show, Eq)``

In this case `BinTree` is the type constructor and `Node` and `Leaf` the data data constructors.

Remember that the goal of folds is to separate the implementation of the recursion from the actual operation we want to execute on the datatype. So we have one function, the fold, that takes care of the recursion and several other functions that use this fold to specify certain semantics on the datatype. We are going to calculate the sum of all elements in this tree.

The above binary tree has elements in the nodes and nothing in the leaves. You can notice the recursive occurrence of ``BinTree a`’ in the node. We see two constructors in this datatype, `Node` and `Leaf`. The Node constructor expects some value of type `a` and two subtrees of type `BinTree a`, and the `Leaf` constructor has no parameters. In Haskell:

``````Node :: a -> BinTree a -> BinTree a -> BinTree a
Leaf :: BinTree a``````

Notice that our list also has two constructors, namely the `(:)` constructor that adds an element and the constructor for the empty list, `[]`. Analogously we have the `Node` and `Leaf` constructors. If we were allowed to use the `(:)` and `[]` constructor in our own Haskell code, the datatype for a list would look like this:

``data List a = (:) a (List a) | []``

Also, it is customary to keep the arguments for the functions in the same order as the constructors are defined in the datatype, and to name the identifiers containing the functions the same as the constructor function but in lowercase. Applying this we get:

``````foldBinTree :: (a -> BinTree a -> BinTree a -> BinTree a) -> (BinTree a) -> BinTree a -> ??
foldBinTree node leaf = f
where f (Node x left right) = node x (f left) (f right)
f (Leaf)              = leaf``````

I left out the result type of this fold, try to find it yourself before continuing.

By following our code we can see that every case, the case for `Node` and the one for `Leaf`, result in a `BinTree a`. Consequently the complete result of the function is a `BinTree a`

Now recall that we previously used a fold to calculate the sum of a list. With that fold we were not restricted to a list. Which is also clearly visible by looking at the type of `foldr` it contains only type variables. As we want to calculate sum of the elements in this tree, which is something of type `a` and not of type `BinTree a` we have to revise the type of our fold. Notice the recurrence of the datatype in it’s declaration, `BinTree a`. We will replace all of these occurrences in the types of our functions with a free type variable, say `r`:

``````foldBinTree :: (a -> r -> r -> r) -> (r) -> BinTree a -> r
foldBinTree node leaf = f
where f (Node x left right) = node x (f left) (f right)
f (Leaf)              = leaf``````

Keep in mind that I used ‘eta reduction’ in the above code. This means that the last parameter (in this case the tree) isn’t explicitly specified as it would appear at the very end of the parameter list and ath the very end of the definition. (More on eta reduction.) So, now let’s do something with this fold. Suppose we have the following tree:

``````bintree = Node 1
(Node 2
(Node 3 Leaf Leaf)
(Node 3 Leaf Leaf)
)
(Node 2
(Node 3 Leaf Leaf)
(Node 3 Leaf Leaf)
)``````

A visual representation:

```            1
/
2    2
/ \  /
3   3 3  3```
Now we can define several traversals over this tree. Let’s calculate the sum of all values in the tree.

``````sumBinTree :: Num a => BinTree a -> a
sumBinTree = foldBinTree (\val res_left res_right -> val + res_left + res_right)
0``````

#### Algebras

The type of our foldBinTree is now more compact. However, there will be datatypes that contain a larger sum of constructors that also may have more or less parameters than in our case. Defining the type of the fold on those datatypes as we have done before will inevitably lead to very clumsy type signatures. The idea is that we split up the section that denotes our functions into a separate type, namely the algebra.

So now we define an algebra for our data structure. To do this we again look at each data constructor and determine it’s type. The fold function takes care of the recursion, so applying this thought consequently gives us the following types and fold.

Keep in mind that we have to change the type of our fold, but not the definition of the fold itself!

``````type BinTreeAlgebra a r = (a -> r -> r -> r, -- Node
r -- Leaf
)

foldBinTree :: BinTreeAlgebra a r -> BinTree a -> r``````

Now we can again define a sum on all `Num a` trees. Notice that we went from to seperate parameters for the functions to one tuple with two elements.

``````sumBinTree :: Num a => BinTree a -> a
sumBinTree = foldBinTree (\val res_left res_right -> val + res_left + res_right,
0)``````
``````\$ ghci bintree.hs
*Main> bintree
Node 1 (Node 2 (Node 3 Leaf Leaf) (Node 3 Leaf Leaf)) (Node 2 (Node 3 Leaf Leaf) (Node 3 Leaf Leaf))
*Main> sumBinTree bintree
19``````

That’s it for now. If you did not understand everything by the end of this article, don’t panic. It will sink in eventually. Let this rest a day or two, and then read this article again and you’ll understand folds better. :)