Transcript Functional Programming - SLU Mathematics and Computer Science

PROGRAMMING IN HASKELL
I/O and functors
Based on lecture notes by Graham Hutton
The book “Learn You a Haskell for Great Good”
(and a few other sources)
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File I/O
So far, we’ve worked mainly at the prompt, and
done very little true input or output. This is logical
in a functional language, since nothing has side
effects!
However, this is a problem with I/O, since the whole
point is to take input (and hence change some
value) and then output something (which requires
changing the state of the screen or other I/O
device.
Luckily, Haskell offers work-arounds that separate
the more imperative I/O.
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A simple example: save the following file as
helloword.hs
main = putStrLn "hello, world"
Now we actually compile a program:
$ ghc --make helloworld
[1 of 1] Compiling Main
( helloworld.hs, helloworld.o )
Linking helloworld ...
$ ./helloworld
hello, world
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What are these functions?
ghci> :t putStrLn
putStrLn :: String -> IO ()
ghci> :t putStrLn "hello, world"
putStrLn "hello, world" :: IO ()
So putStrLn takes a string and returns an I/O action
(which has a result type of (), the empty tuple).
In Haskell, an I/O action is one with a side effect usually either reading or printing. Usually some kind
of a return value, where () is a dummy value for no
return.
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An I/O action will only be performed when you give
it the name “main” and then run the program.
A more interesting example:
main = do
putStrLn "Hello, what's your name?”
name <- getLine
putStrLn ("Hey " ++ name ++ ",
you rock!")
Notice the do statement - more imperative style.
Each step is an I/O action, and these glue together.
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More on getLine:
ghci> :t getLine
getLine :: IO String
This is the first I/O we’ve seen that doesn’t have an
empty tuple type - it has a String.
Once the string is returned, we use the <- to bind
the result to the specified identifier.
Notice this is the first non-functional action we’ve
seen, since this function will NOT have the same
value every time it is run! This is called “impure”
code, and the value name is “tainted”.
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An invalid example:
nameTag = "Hello, my name is " ++ getLine
What’s the problem? Well, ++ requires both
parameters to have the same type.
What is the return type of getLine?
Another word of warning: what does the following
do?
name = getLine
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Just remember that I/O actions are only performed
in a few possible places:
- A main function
- inside a bigger I/O block that we have composed
with a do (and remember that the last action can’t
be bound to a name, since that is the one that is the
return type).
-At the ghci prompt:
ghci> putStrLn "HEEY"
HEEY
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You can use let statements inside do blocks, to call
other functions (and with no “in” part required):
import Data.Char
main = do
putStrLn "What's your first name?"
firstName <- getLine
putStrLn "What's your last name?"
lastName <- getLine
let bigFirstName = map toUpper firstName
bigLastName = map toUpper lastName
putStrLn $ "hey " ++ bigFirstName ++ " " ++
bigLastName ++ ", how are you?"
Note that <- is for I/O, and let for expressions.
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Return in haskell: NOT like other languages.
main = do
line <- getLine
if null line
then return ()
else do
putStrLn $ reverseWords line
main
reverseWords :: String -> String
reverseWords = unwords . map reverse . words
Note: reverseWords = unwords . map reverse .
words is the same as
reverseWords st = nwords (map reverse (words st))
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What is return?
Does NOT signal the end of execution! Return
instead makes an I/O action out of a pure value.
main = do
a <- return "hell"
b <- return "yeah!"
putStrLn $ a ++ " " ++ b
In essence, return is the opposite of <-. Instead of
“unwrapping” I/O Strings, it wraps them.
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Last example was a bit redundant, though – could
use a let instead:
main = do
let a = "hell"
b = "yeah"
putStrLn $ a ++ " " ++ b
Usually, you’ll use return to create I/O actions that
don’t do anything (but you have to have one
anyway, like an if-then-else), or for the last line of a
do block, so it returns some value we want.
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Other I/O functions:
-print (works on any type in show, but calls show
first)
-putStr - And as putStrLn, but no newline
-putChar and getChar
main = do print True
print 2
print "haha"
print 3.2
print [3,4,3]
main = do
c <- getChar
if c /= ' '
then do
putChar c
main
else return ()
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More advanced functionality is available in
Control.Monad:
import Control.Monad
import Data.Char
main = forever $ do
putStr "Give me some input: "
l <- getLine
putStrLn $ map toUpper l
(Will indefinitely ask for input and print it back out
capitalized.)
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Other functions:
• sequence: takes list of I/O actions and does them
one after the other
• mapM: takes a function (which returns an I/O)
and maps it over a list
Others available in Control.Monad:
• when: takes boolean and I/O action. If bool is
true, returns same I/O, and if false, does a return
instead
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System Level programming
Scripting functionality deals with I/O as a necessity.
The module System.Environment has several to help
with this:
• getArgs: returns a list of the arguments that the
program was run with
• getProgName: returns the string which is the
program name
(Note: I’ll be assuming you compile using “ghc –
make myprogram” and then running “./myprogram”.
But you could also do “runhaskell myprogram.hs”.)
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An example:
import System.Environment
import Data.List
main = do
args <- getArgs
progName <- getProgName
putStrLn "The arguments are:"
mapM putStrLn args
putStrLn "The program name is:"
putStrLn progName
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The output:
$ ./arg-test first second w00t "multi word
arg"
The arguments are:
first
second
w00t
multi word arg
The program name is:
arg-test
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Recap of Typeclasses
We have seen typeclasses, which describe classes
of data where operations of a certain type make
sense.
Look more closely:
class Eq
(==)
(/=)
x ==
x /=
a where
:: a ->
:: a ->
y = not
y = not
a -> Bool
a -> Bool
(x /= y)
(x == y)
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Now – say we want to make a new type and make
sure it belongs to a given typeclass. Here’s how:
data TrafficLight = Red | Yellow | Green
instance Eq TrafficLight where
Red == Red = True
Green == Green = True
Yellow == Yellow = True
_ == _ = False
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Now maybe we want to be able to display these at
the prompt. To do this, we need to add this to the
“show” class.
(Remember those weird errors with the trees
yesterday? We hadn’t added trees to this class!)
instance
show
show
show
Show TrafficLight where
Red = "Red light"
Yellow = "Yellow light"
Green = "Green light"
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And finally, we can use these things:
ghci> Red == Red
True
ghci> Red == Yellow
False
ghci> Red `elem` [Red, Yellow, Green]
True
ghci> [Red, Yellow, Green]
[Red light,Yellow light,Green light]
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Functors
Functors are a typeclass, just like Ord, Eq, Show,
and all the others. This one is designed to hold
things that can be mapped over; for example, lists
are part of this typeclass.
class Functor f where
fmap :: (a -> b) -> f a -> f b
This type is interesting - not like previous exmaples,
like in EQ, where (==) :: (Eq a) => a -> a -> Bool.
Here, f is NOT a concrete type, but a type
constructor that takes one parameter.
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Compare fmap to map:
fmap :: (a -> b) -> f a -> f b
map :: (a -> b) -> [a] -> [b]
So map is a lot like a functor! Here, map takes a
function and a list of type a, and returns a list of
type b.
In fact, can define map in terms of fmap:
instance Functor [] where
fmap = map
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Notice what we wrote:
instance Functor [] where
fmap = map
We did NOT write “instance Functor [a] where…”,
since f has to be a type constructor that takes one
type.
Here, [a] is already a concrete type, while [] is a
type constructor that takes one type and can
produce many types, like [Int], [String], [[Int]], etc.
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Another example:
instance Functor Maybe where
fmap f (Just x) = Just (f x)
fmap f Nothing = Nothing
Again, we did NOT write “instance Functor (Maybe
m) where…”, since functor wants a type constructor.
Mentally replace the f’s with Maybe, so fmap acts
like (a -> b) -> Maybe a -> Maybe b.
If we put (Maybe m), would have (a -> b) ->
(Maybe m) a -> (Maybe m) b, which looks wrong.
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Using it:
ghci> fmap (++ " HEY GUYS IM INSIDE THE
JUST") (Just "Something serious.")
Just "Something serious. HEY GUYS IM INSIDE TH
E JUST"
ghci> fmap (++ " HEY GUYS IM INSIDE THE
JUST") Nothing
Nothing
ghci> fmap (*2) (Just 200)
Just 400
ghci> fmap (*2) Nothing
Nothing
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Back to trees, as an example to put it all together:
Let’s make a binary search tree type. Need
comparisons to make sense, so want the type to be
in Eq.
Also going to have it be Show and Read, so anything
in the tree can be converted to a string and printed
(just to make displaying easier).
data Tree a = EmptyTree
| Node a (Tree a) (Tree a)
deriving (Show,Read,Eq)
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Note that this is slightly different than our last class.
Node 5 (Node 3 (Node 1 EmptyTree EmptyTree) (Node 4
EmptyTree EmptyTree)) (Node 6 EmptyTree EmptyTree)
This will let us code an insert that’s a bit easier to
process, though!
First step – a function to make a single node tree:
singleton :: a -> Tree a
singleton x = Node x EmptyTree EmptyTree
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Now – go code insert!
treeInsert :: (Ord a) => a -> Tree a -> Tree a
treeInsert x EmptyTree =
treeInsert x (Node a left right) =
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My insert:
treeInsert :: (Ord a) => a -> Tree a -> Tree a
treeInsert x EmptyTree = singleton x
treeInsert x (Node a left right)
| x == a = Node x left right
| x < a = Node a (treeInsert x left) right
| x > a = Node a left (treeInsert x right)
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Find:
findInTree :: (Ord a) => a -> Tree a -> Bool
findInTree x EmptyTree = False
findInTree x (Node a left right)
| x == a = True
| x < a = findInTree x left
| x > a = findInTree x right
Note: If this is an “unordered” tree, would need to
search both left and right subtrees.
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An example run:
ghci> let nums = [8,6,4,1,7,3,5]
ghci> let numsTree = foldr treeInsert EmptyTre
e nums
ghci> numsTree
Node 5 (Node 3 (Node 1 EmptyTree EmptyTree) (N
ode 4 EmptyTree EmptyTree)) (Node 7 (Node 6 Em
ptyTree EmptyTree) (Node 8 EmptyTree EmptyTre
e))
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Back to functors:
If we looked at fmap as though it were only for
trees, it would look something like:
(a -> b) -> Tree a -> Tree b
We can certainly phrase this as a functor, also:
instance Functor Tree where
fmap f EmptyTree = EmptyTree
fmap f (Node x leftsub rightsub) =
Node (f x) (fmap f leftsub)
(fmap f rightsub)
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Using the tree functor:
ghci> fmap (*2) EmptyTree
EmptyTree
ghci> fmap (*4) (foldr treeInsert
EmptyTree [5,7,3,2,1,7])
Node 28 (Node 4 EmptyTree (Node 8 EmptyTree (N
ode 12 EmptyTree (Node 20 EmptyTree EmptyTree
)))) EmptyTree
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