*Note: this is the completed version of lecture Classes. The lhs version of this file is available.*

# Type Classes

```
> {-# LANGUAGE MultiParamTypeClasses, FlexibleInstances, InstanceSigs #-}
> {-# OPTIONS -fno-warn-type-defaults #-}
> module Classes where
```

```
> import Test.HUnit
> import Data.Char
> import Text.Read
> import Prelude hiding (lookup)
```

# Our first qualified type

Question: What is the type of (+)?

We've most often used (+) to add Ints, as in:

```
> fancySeven :: Int
> fancySeven = 3 + 4
```

So you might guess that the type of (+) is:

`(+) :: Int -> Int -> Int`

But if you think a little harder, you may remember we've also used (+) to add `Float`

s and `Double`

s, as in:

```
> fancyEight :: Float
> fancyEight = 3.2 + 4.8
```

So it must also be the case that:

`(+) :: Float -> Float -> Float`

At this point, you might guess that (+) has the type

`(+) :: a -> a -> a`

since it seems to work for many different types. But this type would be too general: it doesn't make much sense to add a `Bool`

to a `Bool`

or an `Char -> Char`

to an `Char -> Char`

.

We need a type in the middle: (+) should work on any kind of numbers, but not on other things. If we look up the actual type, we find this:

`(+) :: Num a => a -> a -> a`

What's going on here? What's that fancy => thing?

In this type, `Num a`

is a "type class constraint". The type says that (+) should work at any type `a`

, so long as `a`

is a member of the `Num`

type class.

`Num`

is one of many type classes in Haskell's standard library. Types like `Int`

and `Double`

are members of this class because they support operations like `(+)`

and `(-)`

.

Type classes are Haskell's solution for *overloading*, the ability to define functions with the same name but different operations.

For example, the operation of `(+)`

for `Int`

s is very different than that for `Double`

s. The compiler generates different machine instructions!

This is the key difference between overloaded functions, like `(+)`

, and (parametrically)-polymorphic functions, like list `length`

. The `length`

function behaves the same, no matter whether it is working with a list of `Int`

s or a list of `Double`

s. However, the behavior of `(+)`

really does depend on the type of argument that it is working with.

For example, Haskell includes the type `Rational`

to avoid well-known imprecisions in floating point arithmetic. Compare the difference between `x1`

and `x2`

below:

```
> x1 :: Double
> x1 = 1/10 + 2/10
```

```
> x2 :: Double
> x2 = fromRational rat where
> rat :: Rational
> rat = 1/10 + 2/10
```

# Eq

Let's consider another function we've been using quite a bit:

`(==) :: Eq a => a -> a -> Bool`

Again, this makes sense: We've used equality at many different types, but it doesn't work at *every* type: there is no obvious way to check for equality on functions, for example.

Let's peek at the definition of the Eq *type class*:

```
class Eq a where
(==) :: a -> a -> Bool
(/=) :: a -> a -> Bool
(/=) = \c1 c2 -> not (c1 == c2)
```

This declares `Eq`

to be a type class with a single parameter, `a`

. To show that some type is a member of the class, we must provide definitions of `(==)`

and `(/=)`

for the type. We do this with an "instance" declaration.

For example, consider the following type:

`> data PrimaryColor = Red | Green | Blue`

We can tell Haskell that `PrimaryColor`

s can be compared for equality like this:

```
> instance Eq PrimaryColor where
> Red == Red = True
> Blue == Blue = True
> Green == Green = True
> _ == _ = False
```

`> c1 /= c2 = True`

Now we can use (==) and (/=) on `PrimaryColor`

s!

```
> fancyTrue :: Bool
> fancyTrue = Red == Red
```

It might seem annoying, though, that we had to provide both (==) and (/=)...

Fortunately, we don't. Type classes are allowed to provide "default instances". For example, the full definition of `Eq`

from the Prelude is:

```
class Eq a where
(==),(/=) :: a -> a -> Bool
x /= y = not (x == y)
x == y = not (x /= y)
```

So to define `Eq`

for a new type, we only actually have to provide one of (==) and (/=). Haskell can figure out the other for us.

Let's do another example. We'd like to define `Eq`

for the type `Tree`

that we saw last time. But we have a bit of a problem: to check if trees are equal, we'll need to know if the data in each pair of `Branch`

s is equal. Put another way, we'll only be able to compare two `Tree a`

s if `a`

is an instance of `Eq`

.

`> data Tree a = Empty | Branch a (Tree a) (Tree a)`

No worries, Haskell lets us put type class constraints on our instance declarations. See if you can finish this instance for trees.

```
> instance Eq a => Eq (Tree a) where
>
> Empty == Empty = True
> Branch x t1 t2 == Branch y u1 u2 = t1 == u1 && t2 == u2 && x == y
> _ == _ = False
>
```

This code tells Haskell how to compare `Tree a`

s for equality as long as it already knows how to compare `a`

s.

Let's try it out:

```
> tree1, tree2 :: Tree Int
> tree1 = Branch 3 (Branch 2 Empty Empty) (Branch 1 Empty Empty)
> tree2 = Branch 3 Empty Empty
```

```
> testTreeEq :: Test
> testTreeEq = TestList [ "tree1 == tree1" ~: tree1 == tree1 ~?= True,
> "tree1 /= tree2" ~: tree1 == tree2 ~?= False,
> "tree1 /= Empty" ~: tree1 == Empty ~?= False ]
```

# More qualified types

We can now explain the types of a few functions that we glossed over before. Type class constraints don't just appear on the functions defined as members of a type class; they can appear anywhere we want to use such a function. For example the standard library function `lookup`

can be used to find a member of an "association list" pairing keys with their associated values. Let's peek at its implementation:

```
> lookup :: Eq a => a -> [(a,b)] -> Maybe b
> lookup _ [] = Nothing
> lookup a ((a',b):ps) = if a == a' then Just b
> else lookup a ps
```

The idea is that `lookup`

checks to see if the given value appears as the first element of any pair in the list. To implement lookup, we need to use the `(==)`

function to check if we've reached the right pair. So, the type of lookup records that there must be an `Eq`

instance for `a`

; otherwise, the compiler wouldn't have an implementation of `(==)`

for this type.

Try commenting out the type of lookup. What type does Haskell infer?

What about a function that uses lookup, what is its type? See if you can guess the type that GHC infers for this function:

```
> lookupDefault x xs def = case lookup x xs of
> Just y -> y
> Nothing -> def
```

# Other overloaded operations?

Quickly, without looking at the rest of the lecture notes, brainstorm as many *overloaded* operations as you can. What overloaded functions have you seen in other languages? What about from your mathematics classes?

Here are some examples:

- serialization -- converting to a string, binary format, JSON, or other format.
- hashCode -- for indexing into hashtables
- truthiness (bool conversion) --- if null then ... else ...
- other conversions --- toInteger
- iterators -- traversing or visiting data structures.
- comparison -- equality

Overloading is sometimes called *ad hoc* polymorphism, for good reason. Allowing unrelated functions to have the same name can lead to messy, hard-to-understand code and unpredictable behavior.

For example, see this talk by Gary Bernhardt from CodeMesh 2012.

Type classes are Haskell's technology to make ad hoc polymorphism less ad hoc. How do they encourage more principled design?

First, the type class itself means that the types of overloaded functions must follow a common pattern. For example, the type of

`(+)`

states that any instance must take two arguments of the*same*type.`(+) :: Num a => a -> a -> a`

Haskell won't allow you to overload

`(+)`

to work with a`String`

and an`Int`

at the same time, for example.Second, type classes usually come with

*laws*, or specific properties that all instances of the type class are expected to adhere to. For example, all instances of`(==)`

should satisfy reflexivity, symmetry and transitivity. Likewise, all instances of`(+)`

should be commutative and associative.Any time you see a type class, you should ask yourself "what are the laws" that should hold for instances of this class?

Note that there is no way for the Haskell language to enforce these laws when instances are defined; the type checker doesn't know about them. So it is up to programmers to check that they hold (informally) for their instances.

# Deriving

We've now written a couple `Eq`

instances ourselves, and you might guess that most of our future `Eq`

instances will have a very similar structure. They will recurse down datatypes, making sure the two terms use the same constructors and that any subterms are equal. And you'd be right!

To avoid this kind of boilerplate, Haskell provides a nifty mechanism called `deriving`

. When you declare a new datatype, you may ask Haskell to automatically derive an `Eq`

instance of this form. For example, if we wanted equality for the `Shape`

type we saw last time, we could have written:

```
> data Point = Point Double Double
> deriving (Eq)
```

```
> data Shape = Circle Point Float
> | Rectangle Point Point
> deriving (Eq)
```

Haskell can derive an `Eq`

instance as long as it already has one available for any data that appears as arguments to constructors. Since it already knows how to compare `Double`

s, this one works.

It won't always work, though, consider this datatype, which can contain functions on `Int`

s.

```
> data IntFunctions = OneArg (Int -> Int)
> | TwoArg (Int -> Int -> Int)
```

There are no `Eq`

instances for functions (this is a classic example of an undecidable problem!). So, if we added `deriving (Eq)`

to this type, we'd get an error. Try it out!

Of course, not every type class supports this "deriving" mechanism. GHC can derive instances of a handful of classes for us (we'll see a few more today), but for most standard library type classes and any classes you define, you must write the instances out yourself.

# Show and Read

Time for a couple more type classes from the standard library.

Though we haven't talked about it explicitly, we've been using Haskell's printing functions quite a bit. Every time we've run code in ghci and looked at the output, Haskell had to figure out how to convert the data to a String. A few times we've even explicitly used a function called `show`

, which converts a value to a `String`

.

Here is the type of `show`

:

`show :: Show a => a -> String`

Aha, another type class! This says that the function `show`

converts an `a`

to a `String`

, as long as `a`

is a member of the `Show`

class. Let's look at the full definition of this class:

```
class Show a where
show :: a -> String
showsPrec :: Int -> a -> ShowS
showList :: [a] -> ShowS
```

To define an instance of `Show`

, you must implement either `show`

or `showsPrec`

. We've already discussed `show`

, which is a bit like Java's `toString`

. The second function, `showsPrec`

, takes an extra `Int`

argument which can be used to indicate the "precedence" of the printing context - this is useful in some situations to determine, for example, whether parentheses are needed in the output. Its return type, `ShowS`

is used for more efficient printing. For now, you don't need to worry about these details. The third function, `showList`

, exists so that users can specify a special way of printing lists of values, if desired for a given type. Usually, though, the default instance works fine.

By convention, instances of `Show`

should produce valid Haskell expressions (i.e., expressions that can be read by the Haskell parser).

In the other direction, there is a type class called `Read`

. The most primitive function in this class is

`read :: Read a => String -> a`

Notice that the type variable `a`

doesn't appear in any arguments to this function. In general, to use `read`

, you must make sure the type of the output is clear from context or provide it specifically. For example, if you try this in ghci

`ghci> read "3"`

you will get an error: ghci doesn't know whether to interpret the string `"3"`

as an `Int`

or a `Float`

or even a `Bool`

. You can help it, though

```
ghci> read "3" :: Int
3
```

What if you can't read?

```
ghci> read "3" :: Bool
*** Exception: Prelude.read: no parse
```

This exception is irritating, as there isn't much you can do to recover from it. Therefore, I like to use the GHC-specific version of `read`

, from the library `Text.Read`

called

` readMaybe :: Read a => String -> Maybe a`

This (non-partial) version provides a way to recover from a misparse; so is often much more useful.

You can see the details of the `Read`

type class in the standard library and in `Text.Read`

. As one might expect, parsing values is a little more complicated than printing them. One important thing to remember, though, is that the `read`

and `show`

functions should be inverses. So, for example

`read (show 3) :: Int`

should return 3, and

`show (read "3" :: Int)`

should return "3".

This should work for all instances of `show`

and `read`

. For any string `x`

that is readable as a value of type `A`

, i.e. (read `x`

is not an error), it should be the case that

`show (read x :: A) == x`

and, if there is an equality instance for `A`

, we should have:

`read (show a) == a`

Both `Show`

and `Read`

are derivable:

```
> data SadColors = Black | Brown | Grey
> deriving (Eq, Show, Read)
```

Notice that if we type a value into ghci and the corresponding type doesn't have a `Show`

instance, we get an error saying ghci doesn't know how to display the value:

```
ghci> Empty
<interactive>:1:1:
No instance for (Show (Tree a0))
arising from a use of `print'
Possible fix: add an instance declaration for (Show (Tree a0))
In a stmt of an interactive GHCi command: print it
ghci> \x -> (x,x)
<interactive>:1:1:
No instance for (Show (t0 -> (t0, t0)))
arising from a use of `print'
Possible fix:
add an instance declaration for (Show (t0 -> (t0, t0)))
In a stmt of an interactive GHCi command: print it
```

# Type classes vs. OOP

At this point, many of you are probably thinking that type classes are a lot like Java's classes or interfaces. You're right! Both provide a way to describe functions that can be implemented for many types.

There are some important differences, though:

In Java, when you define a new class, you must specify all the interfaces it implements right away. Haskell lets you add a new instance declaration at any time.

This is very convenient: we often define new type classes and want to be able to add instances for types that are already around. We wouldn't want to have to change the standard library just to write a new type class and give it an instance for

`Int`

!Type classes are better integrated in the standard library than Java interfaces. In particular, every object in Java has

`equals`

and`toString`

methods. This leads to some silliness, since not every type of data can sensibly be checked for equality or printed effectively. The result is that calling`equals`

on objects that don't actually implement it may result in a run-time error or a nonsensical result.By contrast, Haskell will warn us

*at compile time*if we try to use`(==)`

on a term that doesn't support it. It's all tracked in the types!Haskell supports

*multi-parameter type classes*and multiple inheritance.In Haskell, classes may require that types be members of an arbitrary number of other classes. For example, you might write a class for "Serializable" data that can be written to a file and demand that Serializable types implement both Read and Show:

`class (Read a, Show a) => Serializable a where toFile :: a -> ByteString ...`

Also, type classes in Haskell may have multiple type arguments. Often it's useful to think of such classes as describing a relationship between types. For example, we can define a class:

`class Convertible a b where convert :: a -> b`

Instances of

`Convertible a b`

show how to convert from one type to another. For example, we can convert from`Char`

s to`Int`

s using Haskell's built in`ord`

function, which gets the ascii code of a character:`instance Convertible Char Int where convert = ord`

Or we can convert from

`Tree`

s to`List`

s using an inorder traversal:`instance Convertible (Tree a) [a] where convert = infixOrder`

Java doesn't have analogues for these features.

(Note that to use multi-parameter type classes, you must give ghc the

`MultiParamTypeClasses`

LANGUAGE pragma.)

# Ord

Let's talk about another type class from the standard library. This one is for comparisons. It is used in the type of (<):

`(<) :: Ord a => a -> a -> Bool`

`Ord`

is a type class for things that can be ordered. Here is its definition:

```
class Eq a => Ord a where
compare :: a -> a -> Ordering
(<), (<=), (>), (>=) :: a -> a -> Bool
max, min :: a -> a -> a
```

Notice that to be a member of the `Ord`

class, a type must also have an implementation of `Eq`

.

Most of these methods we've seen before, so let's talk about the one we haven't:

`compare :: Ord a => a -> a -> Ordering`

This uses a new type from the standard library:

`data Ordering = LT | EQ | GT`

So `compare`

takes two terms and tells us whether the first is less than, equal to, or greater than than the second. Most built in types already have `Ord`

instances (try some examples in ghci).

What properties should instances of `Ord`

satisfy? Write some below:

reflexivity for (<), transitivity for (<) (<) is opposite of (>=), etc. compatibility between `compare`

and (<)

`Ord`

is derivable, like `Eq`

, `Show`

and `Read`

. If you're writing your own `Ord`

instance, you only need to provide `compare`

or `(<=)`

; Haskell can fill in the rest.

The `Ord`

type class shows up all over the standard library. For example, `Data.List`

has a function which sorts lists:

`sort :: Ord a => [a] -> [a]`

As you'd expect, we need to know an ordering on `a`

s in order to sort lists of them!

# Overloading and Syntax

Type classes have been a part of Haskell since the beginning of the language design. Because of that it is integrated into the language syntax in somewhat subtle ways.

For example, it is not just functions that are overloaded. What is the type of `3`

?

```
ghci> :type 3
3 :: Num a => a
```

Literal integers, such as `3`

or `552`

are overloaded in Haskell.

How does this work? The answer lies in the `Num`

type class. There is a lot more in `Num`

besides `(+)`

. Let's take a look!

```
ghci> :i Num
class Num a where
(+) :: a -> a -> a
(-) :: a -> a -> a
(*) :: a -> a -> a
negate :: a -> a
abs :: a -> a
signum :: a -> a
fromInteger :: Integer -> a
-- Defined in 'GHC.Num'
```

The parser converts a literal number to an `Integer`

, but then the `Num`

type class can convert that syntax to any numeric type.

This syntax is convenient, because it allows all numeric types to use the same syntax for constants.

` ghc> 1 :: Double`

and

` ghc> 1 :: Integer`

It also allows expressions like this to work, even though (+) requires both arguments to have the same type.

` ghc> 1 + 2.0`

It can also be (ab)-used for types that aren't traditionally considered numbers. What do you think of this instance?

```
> instance Num a => Num [a] where
> fromInteger = repeat . fromInteger
> (+) = zipWith (+)
> (-) = zipWith (-)
> (*) = zipWith (*)
> negate = map negate
> abs = map abs
> signum = map signum
```

Code that uses this instance looks a bit strange...

```
> seven :: [Int]
> seven = 3 + 4
```

... but is it wrong? Do the operations of (+) and (-) behave like you would expect them to?

# Enum and Bounded

Last week, we observed that we could use the [a..b] list syntax on both `Int`

s and `Char`

s. For example:

```
> tenToThirty :: [Int]
> tenToThirty = [10..30]
```

```
> abcde :: [Char]
> abcde = ['a'..'e']
```

But obviously this syntax can't work on every type. Indeed, it works on only the ones that implement the `Enum`

type class! This class describes sequentially ordered types -- i.e., those that can be enumerated.

```
class Enum a where
succ, pred :: a -> a
toEnum :: Int -> a
fromEnum :: a -> Int
-- These are used in haskell's translation of [n..] and [n..m]
enumFrom :: a -> [a]
enumFromThen :: a -> a -> [a]
enumFromTo :: a -> a -> [a]
enumFromThenTo :: a -> a -> a -> [a]
```

Watch out, though. This syntax can lead to some strange behaviors if you don't know exactly what these functions do for your type. What do you think this expression means?

`> wat = [0.8 .. 10]`

or what about this one?

`> wat' = [0.5 .. 10]`

OK, one more basic type class. Recall that `Int`

isn't arbitrary precision: it represents an actual machine-sized number in your computer. Of course, this varies from machine to machine (64-bit computers have a lot more `Int`

s than 32-bit ones). And Haskell supports a number of other bounded datatypes too -- `Char`

, `Word`

, etc.

It would sure be nice if there were a uniform way to find out how big these things are on a given computer...

Enter `Bounded`

!

```
class Bounded a where
minBound, maxBound :: a
```

So to find the biggest `Int`

on my machine, we can write:

```
> biggestInt :: Int
> biggestInt = maxBound
```

Of course, if we tried to write

```
biggestInteger :: Integer
biggestInteger = maxBound
```

we would get a type error, since `Integer`

s are unbounded. Again the compiler protects us from basic mistakes like this.

Many standard library types support `Enum`

and `Bounded`

. They are also both derivable - but only for datatypes whose constructors don't take any arguments.

# Functor

Now it's time for everybody's favorite type class...

Recall the `map`

function on lists

`map :: (a -> b) -> [a] -> [b]`

that takes a function and applies it to every element of a list, creating a new list with the results. We also saw that the same pattern can be used for `Tree`

s:

`treeMap :: (a -> b) -> Tree a -> Tree b`

If you think a little, you'll realize that `map`

makes sense for pretty much any data structure that holds a single type of values. It would be nice if we could factor this pattern out into a class, to keep track of the types that support `map`

.

Behold, `Functor`

:

```
class Functor f where
fmap :: (a -> b) -> f a -> f b
instance Eq a => Eq (Tree a) where
...
instance Functor Tree where
...
```

`Functor`

is a little different than the other classes we've seen so far. It's a "constructor" class, because the types it works on are constructors like `Tree`

, `Maybe`

and `[]`

-- ones that take another type as their argument. Notice how the `f`

in the class declaration is applied to other types.

The standard library defines:

```
instance Functor [] where
-- fmap :: (a -> b) -> [a] -> [b]
fmap = map
```

We can define a `Functor`

instance for our own trees:

```
> instance Functor Tree where
> fmap = treeMap where
> treeMap f Empty = Empty
> treeMap f (Branch x l r) = Branch (f x) (treeMap f l) (treeMap f r)
```

The standard library also defines `Functor`

instances for a number of other types. For example, `Maybe`

is a `Functor`

:

```
instance Functor Maybe where
fmap f (Just x) = Just (f x)
fmap f Nothing = Nothing
```

`Functor`

is very useful, and you'll see many more examples of it in the weeks to come.

See if you can define a Functor instance for this type:

`> data Two a = MkTwo a a deriving (Eq, Show, Read, Ord)`

```
> instance Functor Two where
>
> fmap :: (a -> b) -> Two a -> Two b
> fmap f (MkTwo x1 x2) = MkTwo (f x1) (f x2)
```

In the meantime, think about what laws instances of this class should satisfy. (We'll come back to this.)

# Monad

Last, the most famous of all Haskell type classes: The warm fuzzy thing called 'Monad'.

We saw an example of the IO monad with code like this:

```
> main :: IO ()
> main = do
> putStrLn "This is the Classes lecture. What is your name?"
> inpStr <- getLine
> putStrLn $ "Welcome to Haskell, " ++ inpStr ++ "!"
> putStrLn $ "Now running tests."
> _ <- runTestTT testTreeEq
> return ()
```

This code works because `IO`

is an instance of the `Monad`

type class. We'll see more instances of this class in the next few lectures. Don't try to understand it all at once! We'll start with just looking at what's going on at a syntactic level.

```
class Monad m where
-- | Sequentially compose two actions, passing any value produced
-- by the first as an argument to the second.
(>>=) :: m a -> (a -> m b) -> m b
-- | Inject a value into the monadic type.
return :: a -> m a
-- | Sequentially compose two actions, discarding any value produced
-- by the first, like sequencing operators (such as the semicolon)
-- in imperative languages.
(>>) :: m a -> m b -> m b
m >> k = m >>= \_ -> k -- default definition
-- | Fail with a message. This operation is not part of the
-- mathematical definition of a monad, but is invoked on pattern-match
-- failure in a @do@ expression.
fail :: String -> m a
fail s = error s -- default definition
```

You can see the use of `return`

in the last line of the `main`

function above. In fact, it must be the last line in any computation because it doesn't compose multiple actions together.

```
> nop :: IO ()
> nop = do
> return ()
```

We've also been using `(>>=)`

, but only behind the scenes. You've missed it because of another feature of Haskell---the "do" syntax for composing sequences of actions.

For example, code like this

```
main :: IO ()
main = do
x <- doSomething
doSomethingElse
y <- andSoOn
return ()
```

is really shorthand for this:

```
doSomething >>= ( \x ->
doSomethingElse >>
(andSoOn >>= ( \y ->
return () )))
```

So everytime that you see `do`

there is some monad involved (though, as we'll see later, not necessarily `IO`

!).