*(This is an edited repost of one of the posts from the earlier*

version of my Haskell tutorial.)*(This file is a literate haskell script. If you save it*

as a file whose name ends in ".lhs", it's actually loadable and

runnable in GHCI or Hugs.)

Like any other modern programming language, Haskell has excellent support

for building user-defined data types. In fact, even though Haskell is very

much *not* object-oriented, most Haskell programs end up being centered

around the design and implementation of data structures, using constructions

called *classes* and *instances*.

In this post, we're going to start looking at how you implement data types

in Haskell. What I'm going to do is start by showing you how to implement a

simple binary search tree. I'll start with a very simple version, and then

build up from there.

There are a variety of useful primitive types in Haskell - a wide range of

numeric types (arbitrary sized integers, rationals, floats, etc.), characters,

and so on. There are also a lot of basic compound types provided by the

standard library - the usual tuples, arrays, lists, and so on. But when you go

to build your own datatypes, the main thing that you use in Haskell is

something called an *algebraic data types*. An algebraic data type

consists of a set of *variants*, each of which is (essentially) a tuple

which has been tagged with a marker that identifies which variant it belongs

to. When you define an algebraic type, Haskell creates a set of

*constructor functions* for the type, one for each variant.

As usual, talking about things like this in theory is hopelessly vague, so

let's dive in and look at some code. We'll start with a very simple binary

search tree containing integers.

The basics are pretty straightforward. A binary search tree is a data

structure consisting of a collection of nodes. Every node contains a value, a

left child, and a right child. The left and right children each contain

another binary search tree which could contain some values, or which could be

empty.

When you're describing a data type and you use the word "or", that's your

clue that you're looking at a variant. The tree type has two cases: trees

containing values, and empty trees. A declaration of that type would look

like:

>data SimpleIntBST = SIBNode Integer SimpleIntBST SimpleIntBST > | SIBEmpty

What this says is: we're defining a data type named

`SimpleIntBST`

. There are two kinds of `SimpleIntBST`

values. The first kind is an `SIBNode`

, which contains an integer

value, a left child, and a right child. The other kind is an

`SIBEmpty`

, which contains no values. In a language

like C++ or Java, you'd have each node contain pointers to children

nodes, and an empty child would be represented by a null pointer. But

in Haskell, there are no pointers - and more importantly, null doesn't

have a well-defined type. In Haskell, when you've got something

like a null, you need to explicity define the null case using

a variant, like `SIBEmpty`

.

In the declaration, `SIBNode`

and `SIBEmpty`

are the

names of the *constructor functions* that are used to create values. To

create an `SIBNode`

, you call it like a normal function with

parameters of the correct type: for example, to create an `SIBNode`

containing the value "4" with empty left and right children, you'd write:

SIBNode 4 SIBEmpty SIBEmpty

Let's try that in GHCI:

*Main> SIBEmpty <interactive>:1:0: No instance for (Show SimpleIntBST) arising from use of `print' at <interactive>:1:0-7 Possible fix: add an instance declaration for (Show SimpleIntBST) In the expression: print it In a 'do' expression: print it

What's wrong? The interpreter tried to print out the value returned by the

expression. The interpreter outputs values using the `print`

function. In general in Haskell, when we have a problem like this, we look at

types. What's the type of the `print`

function?

*Main> :type print print :: (Show a) => a -> IO () *Main>

The parameter to `print`

must be a value of some type that is

an instance of the `Show`

class. But we didn't make our binary tree

an instance of `Show`

. Fortunately, since `Show`

is such

a fundamental interface, Haskell provides a nifty shorthand to make the

compiler *generate* the code necessary to provide an implementation of

whatever is needed for `Show`

, called a *deriving* clause. Let's

try again, this time with a deriving clause. We'll call the new version

"IntBST" rather than "SimpleIntBST", because treating this post as a literate

file, we can't re-declare "SimpleIntBST".

>data IntBST = IBNode Integer IntBST IntBST > | IBEmpty deriving (Show)

Now, loading that into GHC, we'll see:

*Main> IBNode 4 IBEmpty IBEmpty IBNode 4 IBEmpty IBEmpty *Main>

Which is exactly what we want: our binary search tree is now printable.

A data structure like this isn't useful without some operations to work

with it. So let's go ahead and do some implementing! We'll start with an

insert operator:

>ibstInsert :: IntBST -> Integer -> IntBST >ibstInsert (IBNode i left right) v > | i > v = IBNode i (ibstInsert left v) right > | otherwise = IBNode i left (ibstInsert right v) >ibstInsert IBEmpty v = (IBNode v IBEmpty IBEmpty)

So let's look at that a bit. It starts off with an explicit type

declaration. That's not strictly necessary: the compiler can infer the type of

the function without any help. But it's a good practice to put it there: it's

a useful bit of documentation for a human reader, and it's useful for

catching problems during compilation - if you made any mistakes writing

the function, the compiler might be able to catch them because the

inferred type won't match the declaration.

After the type declaration, we get to the real definition, which is where

things get interesting. I'm sure you noticed that there was something a bit

strange about the type declaration for `IntBST`

. In most

non-functional languages, you define a new data type as some kind of a

*record* or *structure*, which contains a list of *named*

fields; when you use the structure, you access its fields by derefencing an

instance of the data structure using the name of a field. But in our Haskell

binary search tree, the values belonging to an instance of the data structure

*don't have names*. Like most functional languages, Haskell uses

*pattern matching* to access the pieces of a data structure.

A *pattern* looks like a value expression, except that it may

contain some *unbound* variables. The pattern can also be followed by a

*guard expression* which is a boolean expression using variables bound

in the pattern. If the guard is included, it's separated from the patterns by

a `|`

character, which is generally read as "such that". So the

first declaration line of `ibstInsert`

has two patterns:

`(IBNode i left right)`

, and `v`

. It's followed by a

guard. So that full declaration line can be read as "ibstInsert of an IntBST

matching '`(IBNode i left right)`

' and a value matching

'`v`

' *such that* '`i > v`

' returns

'`(IBNode i left right)`

'.

As a shorthand, if you have a guard expression on an indented line

following a pattern with a guard, the pattern is reused with the second guard.

So the second line starting with a "`|`

" is used if the first guard

fails. As a guard expression, "`otherwise`

" can only be used as the

*last* guard for a pattern, and it's always true.

So the first case of the declaration of `ibstInsert`

, in full,

does the following:

- If the tree parameter is an
`IBNode`

, then bind "i" to

the value element of the node, "left" to the left subtree, and "right"

to the right subtree; and bind the variable "v" to the second parameter,

which is the value to be inserted to the tree. - If the value in the tree node is larger than the value being

inserted, then return a tree that's the same as this, except that

the value is inserted into the*left*subtree. - Otherwise, return a tree like this one, except that the new

value is inserted into the*right*subtree.

The second case says that if the node is an empty, then return a new

non-empty node with the new value as its value, and with empty left and right

children.

If you look at the code now that you know how to read it, one thing that

should hopefully be rather striking about it is just how much it looks like

the explanatory pseudo-code that you'd find in an algorithms textbook. It's

very straightforward code, written in a very easy to read style. (As a brief

aside, this is an example of one of the things that really attracts me to

Haskell programming. Code in Haskell tends to very naturally decompose into

very small definitions that look and read like explanatory code from a

textbook.)

So let's look at a quick example of using this to build a tree.

> treeone = ibstInsert (ibstInsert (ibstInsert (ibstInsert (ibstInsert IBEmpty 3) 5) 2) 17) 10 > treetwo = (ibstInsert (ibstInsert (ibstInsert (ibstInsert treeone 4) 6) 1) 9)

That gives us two trees, one of which contains a subset of the values in

the other. We can ask ghc to show us the values of them:

*Main> treeone IBNode 3 (IBNode 2 IBEmpty IBEmpty) (IBNode 5 IBEmpty (IBNode 17 (IBNode 10 IBEmpty IBEmpty) IBEmpty)) *Main*gt; treetwo IBNode 3 (IBNode 2 (IBNode 1 IBEmpty IBEmpty) IBEmpty) (IBNode 5 (IBNode 4 IBEmpty IBEmpty) (IBNode 17 (IBNode 10 (IBNode 6 IBEmpty (IBNode 9 IBEmpty IBEmpty)) IBEmpty) IBEmpty)) *Main>

The other thing that we want to be able to do with a binary search tree is

find nodes in it. Since our current BST only contains integers, we'll just

write a function that returns a boolean value indicating whether or not a

particular value is in the tree.

> >ibstSearch :: IntBST -> Integer -> Bool >ibstSearch (IBNode k left right) v > | k == v = True > | k > v = ibstSearch left v > | otherwise = ibstSearch right v >ibstSearch IBEmpty _ = False >

This one should be pretty easy to read without much explanation. To be

pedantic, I'll walk through it quickly. If it's a non-empty node, and the

value in the node equals the value being searched for, return true. If the

value in the node is bigger, then return the result of searching the left

subtree; otherwise, return the result of searching the right subtree. If it's

an empty node, then the value isn't there, so return false.

A binary search tree isn't particularly useful if it can only hold

integers. So as a first step towards making it more real, we'd like to be able

to store not just integers in the tree, but any value for which the necessary

comparisons (`>`

and `==`

) are defined. There's a

typeclass for that, called "`Ord`

", for *Ordered*. So now

let's just rewrite `IntBST`

as a generic BST using typeclasses:

> data (Ord a) => GenBST a = GBNode a (GenBST a) (GenBST a) > | GBEmpty > deriving (Show)

The variable on the *left* of the equal in the data type definition

is a type parameter. The type-class constraint preceding the name of the new

datatype means that the type parameter "`a`

" can *only* be

instantiated by a type that is an instance of the type-class

"`Ord`

". The method implementations will barely change:

> gbstInsert :: (Ord a) => GenBST a -> a -> GenBST a > gbstInsert (GBNode i left right) v > | i > v = GBNode i (gbstInsert left v) right > | otherwise = GBNode i left (gbstInsert right v) > gbstInsert GBEmpty v = (GBNode v GBEmpty GBEmpty) > > gbstSearch :: (Ord a) => GenBST a -> a -> Bool > gbstSearch (GBNode k left right) v > | k == v = True > | k > v = gbstSearch left v > | otherwise = gbstSearch right v > gbstSearch GBEmpty _ = False

With that implementation of `gbstInsert`

, we can pretty much

reuse the code to generate a sample instance:

> gtreeone = gbstInsert (gbstInsert > (gbstInsert (gbstInsert (gbstInsert GBEmpty 3) 5) 2) 17) 10 > gtreetwo = (gbstInsert (gbstInsert (gbstInsert (gbstInsert gtreeone 4) 6) 1) 9)

Look what ghci says about those:

*Main*gt; gtreetwo GBNode 3 (GBNode 2 (GBNode 1 GBEmpty GBEmpty) GBEmpty) (GBNode 5 (GBNode 4 GBEmpty GBEmpty) (GBNode 17 (GBNode 10 (GBNode 6 GBEmpty (GBNode 9 GBEmpty GBEmpty)) GBEmpty) GBEmpty)) *Main> :type gtreetwo gtreetwo :: GenBST Integer *Main> gbstSearch gtreetwo 17 True *Main> gbstSearch gtreetwo 18 False *Main> gbstSearch gtreetwo 1 True *Main> gbstSearch gtreeone 1 False *Main>

We can use the new generic tree in *exactly* the same way that we

used the integer specific one. We don't need to declare types of expressions

that use it. The compiler will happily infer the type parameter, and just make

everything work. But now we're not restricted to integers:

*Main> gbstInsert (gbstInsert (gbstInsert GBEmpty "hi") "hello") "Salutations" GBNode "hi" (GBNode "hello" (GBNode "Salutations" GBEmpty GBEmpty) GBEmpty) GBEmpty *Main> gbstInsert (gbstInsert (gbstInsert (gbstInsert GBEmpty "hi") "hello") "S alutations") "Greetings" GBNode "hi" (GBNode "hello" (GBNode "Salutations" (GBNode "Greetings" GBEmpty GBEmpty) GBEmpty) GBEmpty) GBEmpty *Main>

That's some of the basics. Next post, we'll build on the binary search

tree, giving it some more features, and updating the insert routine so that

the tree stays nicely balanced. That will lead us in to some interesting

things about how you code in a language like Haskell, where you can't actually

modify data structures in-place.

Mark, can you load and run this in your interpreter? I haven't spent a lot of time on it, but there are at least a few syntax issues I run into when trying to get it to run with GHC 6.8.2 between extraneous spaces and missing blank lines between code and comment.

Dear phlyingpenguin:

Browsers -- or browser/OS pairs -- frequently disagree on what to offer to the cut-and-paste machinery, some omitting the blank line between a code block and the following text, making a hash of otherwise sound literate blogging. (In this case, Safari was wrong for me, Firefox right; but there is the further problem that here the html is also missing some of the spaces between the opening '>'s of the code lines.) I think this paste will be cut-able:

http://hpaste.org/fastcgi/hpaste.fcgi/view?id=13329

Indeed, I keep forgetting that I've switched between Chrome and Safari which both exhibit the same behavior. Firefox does appear to render the way you intend. I had fixed the copy/paste manually without too much more difficulty later on. Thanks!

Dear Mark,

I have a Haskell program to parsing and evaluating an expression. I have the defined the property to perform quickcheck and thus an instance for the Expr which is as follows

instance Arbitrary Expr where

arbitrary = sized arbExpr

arbExpr :: Int -> Gen Expr

arbExpr s =

frequency [ (1, do n