undefined
.
Eventually, the complete
version will be made 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.
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
> t1 == t2 = undefined
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?
FILL IN EXAMPLES HERE
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.
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 aString
and anInt
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
andtoString
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 callingequals
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 fromChar
s toInt
s using Haskell's built inord
function, which gets the ascii code of a character:instance Convertible Char Int where convert = ord
Or we can convert from
Tree
s toList
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:
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 = undefined
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
!).