Transpose a matrix
            transpose::[[a]] -> [[a]]
transpose [] = repeat []
transpose (x:cx) = zipWith(:) x $transpose cx transpose [[1, 2] [3, 4]] zipWith(:) [1, 2]$ transpose [[3, 4]]
[[3],[4] <= zipWith(:) [3, 4] $transpose [[], []..] zipWith(:) [1, 2] [[3], [4]] => [[1, 3], [2, 4]]  The difference among of data, type and newtype Compare data, type and newtype in Haskell Show Java example and C struct, typedef example  data MyClass = Type1 | Type2 String | Type3 Int class MyClass{ public MyClass(){} public MyClass(String s){} public MyClass(Integer s){} } type Str = String // typedef in C, In reality, C does not have String typedef Str String; // typedef in C typedef struct Persons{ char[] name; int age; } Person; Person p; There is not type alias in Java unfortunately newtype NewType a = NewType{name::String -> a}  Haskell Array  import Data.Array import Data.Text.TDFA let ar = array(0, 10) [(i, i*i) | i <- [0..10]] ar ! 1 => 1  Factorial Function in Haskell  fac::Integer -> Integer fac 0 = 1 fac n = n * fac (n - 1) fac n = if n == 0 then 1 else n*fac (n - 1)  Find all prime number in Haskell  prime = sieve[2..] where sieve(p:xs) = p:sieve[ x | x <- xs, mod x p /= 0] -- sieve[2, 3, 4, 5, 6] -- 2:sieve([3, 5]) -- 2:3:sieve([5]) -- 2:3:5:sieve([])  Split String in Haskell  break(\x -> isDigit x == False) "123dog" -- ("123", "dog") break(\x -> isSpace x) "123 dog" -- ("123", " dog") filter(\x -> isDigit x) "123dog" -- "123" let reg = mkRegex "[[:space:]]+" splitRegex r "Agentina Switzerland Sweden" -- ["Agentia", "Switzerland", "Sweden"] -- split string at a particular position splitAt 3 "Agentina" -- ["Age", "ntina"]  Haskell class and instance  Haskell class and instance are very similar to template/generic, interface in Java/C++ 1. Create interface with generic 2. Implement it with concrete type   class Test a where tme::a -> a -> Bool data Dump = Dump{int::Int, str::String} instance Test Dump where tme d1 d2 = (int d1 == int d2) && (str d1 == str d2) -- test it let d1 = Dump{int=3, str="dog"} let d2 = Dump{int=3, str="dog"} print$ tme d1 d2   -- True

// Java Interface with implementation
interface ITest{
public T fun(T t1, T t2);
}

class MyTest implements ITest{
public MyTest(){}
public Integer fun(Integer t1, Integer t2){
return t1 + t2;
}
}

Functor is like a box contains something inside such as Maybe or [1, 2].
A function called fmap which takes a function and abox, fmap applies the function to the value inside thebox and return abox contains the result.

Functor can be formally defined by a pair of functions $f_1 \mbox{ and } f_0$
so that $f_0:Ob(Hask) \rightarrow Ob(Hask)$ and $f_1:Hom(Hask) \rightarrow Hom(Hask)$
where $Hom(Hask)$ refers to the union of all sets $a \rightarrow b$ where $a, b \in Ob(Hask)$
so that the following holds

0. $id::a \rightarrow a$
1. If $g::a \rightarrow b$ then $f_1(g)::f_0(a) \rightarrow f_0(b)$
2. For all $a \in \mbox{Ob(Hask)}, \quad f_1(id_a) = id_{f_0(a)}$
3. If $g,h \in \mbox{Hom(Hask)} \mbox{ then } \quad f_1(g \circ h) = f_1(g) \circ f_1(h)$

Given $f_0(a) = \mbox{List } a \quad f_1(g) = \mbox{ map } g$

Prove $f_0$ and $f_1$ is a Functor
\begin{aligned} &f_0(a) \Rightarrow \mbox{ List } a \quad f_0(b) \Rightarrow \mbox{ List } b \quad \mbox{ [from 1] } \\ &\Rightarrow \quad f_1(g):: \mbox{ List }a \rightarrow \mbox{ List }b \\ &\because \quad id_{f_0(a)}:: f_0(a) \rightarrow f_0(a) \\ &\therefore \quad id_{f_0(a)}:: \mbox{ List }a \rightarrow \mbox{ List }a \\ &\therefore \quad f_1(id_a):: \mbox{ List }a \rightarrow \mbox{ List }a \quad [\mbox{ from 1 } , \mbox{ sub } g::a \rightarrow b \mbox{ with } id::a \rightarrow a] \\ &\Rightarrow f_1(id_a) = id_{f_0(a)} \\~\\ &f_1( g \circ h) \Rightarrow \mbox{ map } g \circ h \Rightarrow \mbox{ map } g \circ \mbox{ map } h \quad \mbox{ [from def. of map]} \\ &f_1(g) \circ f_1(h) \Rightarrow \mbox{ map } g \circ \mbox{ map } h\\ &\Rightarrow f_1( g \circ h) = f_1(g) \circ f_1(h)\\ &\Rightarrow f_0 \mbox{ and } f_1 \mbox{ is a Functor} \\ \end{aligned}
Picture of Functor

Not everything is Functor
MyMaybe is not a Functor since it does't satify the SECOND rule given from above
            data MyMaybe a = MNothing | MyJust Int a deriving(Show)

instance Functor MyMaybe where
fmap f MNothing = MNothing
fmap f (MyJust counter a) = MyJust (counter + 1)  (f a)

main = do
print "Hello World"
print $fmap (+1) MNothing print$ fmap (+1) (MyJust 0 2)
print $fmap (+1)$ fmap (+1) (MyJust 0 2)
print $fmap id (MyJust 0 2) print$ id (MyJust 0 2)
-- fmap id (MyJust 0 2) != id (MyJust 0 2)
-- MyMaybe is not a Functor

Haskell separates pure functions from computations where side effects must be considered by encoding those side effects as values of a particular type. Specifically, a values of type [IO a] is an action, which if executed would product a value type of a
[IO a] is an action
            getLine::IO String
putStrLn::String-> IO()
randomRIO::(Random a) => (a, a) -> IO a

Some IO goodness
            -- define a function to print out "Hello World"
fun::IO()
fun = putStrLn "Hello World"

-- IO action is like a type.
putStrLn::String->IO ()  -- input is String, output is IO()

fun::String->String      -- input is String, output is String

Chain the IO actions
            (>>)::IO a -> IO b -> IO b
where if x and y are IO actions, then (x >> y) is the action performs x, dropping the result
and then performs y and return its result. Great, we can now write programs which do multiple things

main = putStrLn "Hello" >> putStrLn "World"

How to use the result of first action
            (>>=)::IO a -> (a -> IO b) -> IO b
Now, (x >>= y) is the action first perform x, and capture its result, passing the result to f,
which then computes a second actions to be performed. The action is then carried out,
and its result is the result of overall computation.

It does seem to make sense, but what is the concrete example
The definition of readFile and print are as following
            readFile::FilePath->IO String
print::String->IO ()
-- Chain readFile to print with >>=
readFile filePath >>=\file -> print file


    Definition of const
const::a -> b -> a  [keep the first argument and throw away the second argument]
It does't look that good from outside but some inside and be surprised [surprisedly]

const 3 4 -- output 3

>> can be defined in term of >>=
x >> y == x >>= const y
or
x >> y == x >>= \_ -> y

How to turn a value to an IO action which do nothing
return:: a -> IO a

Use IO in function

print contents

IO are first class
create control structure with IO

repeat'::Int->IO () -> IO ()
repeat' n a = a >> repeat'(n-1) a

repeat' 3 putStr('x')   -- output xxx

create for loop
for::[a]->(a -> IO()) -> IO ()
for [] f = return ()
for (x:xs) f = f x >> for xs f

for [1..10] (\x -> putStrLn x)

for [1..10] (\x -> do
putStrLn x
putStrLn x
)

Similarly
forM::(Traversable t, Monad m)=> t a -> (a -> m b) -> t(m b)

let s = ["dog", "cat"]
forM s $\x -> print x dog cat [(), ()]   -- get two Chars getTwoChars::IO (Char, Char) getTwoChars = getChar >>= \c1-> getChar >>= \c2 -> return (c1, c2) -- do statement getTwoCharsDo::IO(Char, Char) getTwoCharsDo = do { c1 <- getChar; c2 <- getChar; return (c1, c2) } -- get line with nested do getMyLine::IO [Char] getMyLine = do { x <- getChar; if x == '\n' then return [] else do{ xs <- getMyLine; return (x:xs) } }  Three weird primitives for composing and chaining together IO actions (>>)::IO a -> IO b -> IO b [kill the first action's result and return the second action's result] (>>=)::IO a -> (a -> IO b) -> IO b return::a -> IO a [wrap a with IO and return it] class Monad a where (>>=)::IO a ->(a -> IO b) -> IO b return::IO a -> (a -> IO b) -> IO b class Monad m where (>>=)::m a ->(a -> m b) -> m b return::a -> m a fun::[(Integer, Integer)] fun n = do x <- [-10..10] if x > 0 then [] else return (n, n+1) Monad [] is the instance of Monad Weird Haskell const  Definition of const in GHC const::a -> b -> a [keep the first argument and throw away the second argument] const 1 2 -- output 1 map(const 3) [1..4] -- output [3, 3, 3] Implement repeat function repeat'::Int->a->[a] repeat' n a = map(const a) [1..n] repeat' 3 "x" -- output ["x", "x", "x"] Implement length function let length' = foldr(\_ acc -> 1 + acc) 0 let length' = foldr(const(1+)) 0 Three weird ways to define >> (>>)::IO a -> IO b -> IO b [this is normal] x >> y = x >>= const y [const is here to save you] x >> y = x >>= \_ -> y [lambda! Ok, it does make sense] x >> y = (. const) . (>>=) [that doesn't make sense for me at least]  let binder inside the map  // let .. in .. => let something then return something let list = [1..10] map(\x -> let square = x*x in square mod 3 == 0) list -- let binding in Haskell -- ax^2 + bx + c = 0 -- Find the root of the quadratic equation roots a b c = let discriminant= sqrt(b^2 - 4*a*c) in ( (-b + discriminant)/(2*a), (-b - discriminant)/(2*a)) main = do print (roots 3 9 1)  Haskell Binary Tree -- binary tree in Haskell data Tree a = Empty | Node a (Tree a) (Tree a) deriving (Show) binsert::Tree Integer->Tree Integer->Tree Integer binsert Empty (Node a Empty Empty) = (Node a Empty Empty) binsert (Node a l r) (Node b Empty Empty) = if b < a then Node a (binsert l (Node b Empty Empty)) r else Node a l (binsert r (Node b Empty Empty)) -- build Binary Tree from preorder and inorder buildTree::[Char]->[Char]->Tree Char buildTree _ [] = Empty buildTree [] _ = Empty buildTree preorder inorder = Node h (buildTree leftPre leftIn) (buildTree rightPre rightIn) where h = head preorder leftIn = filter(\x->x < h) inorder rightIn = filter(\x->x > h) inorder leftPre = take (length rightIn)$ tail preorder
rightPre = subList preorder (length leftIn) $length preorder -- In order traversal in Binary Tree inorder::Tree a->[a] inorder Empty = [] inorder (Node a l r) = (inorder l) ++ [a] ++ (inorder r) -- check whether a Binary Tree is symmetric sym::Tree a ->Bool sym Empty = True sym (Node a Empty Empty) = True sym (Node a l Empty) = False sym (Node a Empty r) = False sym (Node a l r) = sym l && sym r -- find the maximum number of nodes in a Binary Tree maxlen::Tree a->Integer maxlen Empty = 0 maxlen (Node a l r) = 1 + max (maxlen l) (maxlen r) -- find the maximum height of a Binary Tree maxheight = maxlen - 1  What is difference between Expr and Expr a It is hard to figure out the difference between Expr and Expr a in Haskell data Expr is type Int only data Expr a can be any type as 'a'  data Expr = Expr Int | Add Expr Expr | Mul Expr Expr data Expr a = Con a | Add (Expr a) (Expr a) | Mul (Expr a) (Expr a)  If you implement Monoid in the modern GHC, you need to implement Semigroup Semigroup$(\otimes, S)$: let$a, b, c, d \in S$$a \otimes b \otimes c = a \otimes (b \otimes c) \in S$ Monoid$(\otimes, S, I)$: let$a, b, c \in S\begin{aligned} a \otimes b \otimes c &= a \otimes (b \otimes c) \in S \\ I \otimes a &= a \otimes I \end{aligned} Group(\otimes, S, I)$let$a, b, c \in S\begin{aligned} a \otimes b \otimes c &= a \otimes (b \otimes c) \in S \\ I \otimes a &= a \otimes I \\ \forall a \in S, \exists a^{-1} \mid a \otimes a^{-1} &= I \\ \end{aligned}  newtype Sum = Sum{getSum::Int} where deriving (Show) class Semigroup a where (<>):: a -> a -> a instance Semigroup Sum where (<>)::(Sum x) (Sum y) = Sum (x + y) instance Monoid Sum where mappend = (<>) mempty = Sum 0 foldr((<>)) (Sum 0) [(Sum 1), (Sum 2)] => Sum{getSum = 3}  Why Haskell Record is so BAD data Person = Person{ name::String, age::Int } data Dog = Dog{ name::String, age:Int } -- Record is just like function name::Person -> String age::Person -> Int let rec = Person {name ="David", age = 3} name rec -- return David age rec -- return 3  You will get compiler error with two differnet records From above two records: Person and Dog. They both have name::Text In Haskell, name is just a function, it is global. This is why you can NOT have the same name and in different records.  Haskell record is similar to Java Class without methods. You can not have function inside record in Haskell. -- | If you are Java or C++ guys, you might do something like that: -- | There are lots of typing without doing anything useful here:) Java needs 23 lines of code Haskell needs 4 lines of code Of couse there are Setter and Getter generator like Lombok to do that work in Java but you need to import the libraries and manage the dependency.   public class{ String firstName; String lastName; Int age; public String getFirstName(){ return firstName; } public String getLastName(){ return lastName; } public Int getAge(){ return age; } public void setFirstName(String firstName){ this.firstName = firstName; } public void setLastName(String lastName){ this.lastName = lastName; } public void setAge(Int age){ this.age = age; } } -- | /Users/cat/myfile/bitbucket/haskell/Record.hs -- -- | create record called Person with three fields: -- | firstName, lastName and age data Person = Person { firstName::String, lastName::String, age::Int} deriving (Eq) -- | Set some values for the Person, it is very simple, -- | You do not need Setter. -- | -------------------------------------------------------------------------------- let person = Person{firstName = "Jonathan", lastName = "Michell", age = 20} let animal = Person{firstName = "Nonhuman", lastName = "Michell", age = 20} -- | How to access the fields, there are Getters for us by default -- | -------------------------------------------------------------------------------- pp "firstName=" ++ firstName person  -- Jonathan
pp $"lastName =" ++ lastName person -- Michell pp$ "age      =" ++ show(age person)  -- 20
-- | --------------------------------------------------------------------------------

Haskell Record with IORef

IORef is tricky to use initially if you are from Java or C++, If you combine Record with IORef
then it is harder to figure out how the syntax work
1. We want to create a Counter with two fields, both fields are Int type, nothing is revolutionary
2. Increate each fields by certain number
3. Show the certain field


Create Record with two fields:
1. Pos::IORef Int => $Pos \in \{IORef Int\}$
2. Up::IORef Int => $Up \in \{IORef Int\}$
data Count = Count{
pos :: IORef Int,
up :: IORef Int
}

1. Create Counter
2. Create incremental functions
3. Create the show function

    makeCounter::Int->Int-> IO Count
makeCounter c u = do
c' <- newIORef c
u' <- newIORef u
return (Count c' u')

intPosCounter::Int->Count->IO()
intPosCounter n (Count c _) = modifyIORef c (+n)

intUpCounter::Int->Count->IO()
intUpCounter n (Count _ u) = modifyIORef u (+n)

showCounter::Count->IO()
showCounter (Count c u) = do
pp (show c' ++ " " ++ show u')


What does do-statement do in Haskell

do-statement is syntax sugar for Monad Computation??

    readFileToList path >>= \x -> reverse x
do {
let s = reverse s
}

    nameDo::IO String
nameDo = do putStr "What is your first name: "
first <- getLine
putStr "What is your last name: "
last  <- getLine
let full = first ++ " " ++ last
putStrLn("full name " ++ full ++ " ")
return full

main = nameDo >>= \name -> print name

-- alternative
main = do
name <- nameDo
print name

How liftM can help you to simplify your IO code

liftM defintion:
liftM::(Monad m)=>(a -> b) -> m a -> m b

not::Bool -> Bool
not true = false
not false = true

liftM not
liftM::(Monad m)=>(Bool -> Bool) -> m Bool -> m Bool


Real world example with liftM

-- find all the file that does not exist from a list
filterM(\x -> doesFileExist x >>= \y -> return y == false) ["f1.x", "f2.x"]

-- if we find all the files DOES exist in the list, it is easy
filterM(doesFileExist) ["f1.x", "f2.x"]

-- better way to do with liftM
filterM((liftM not) . doesFileExist) ["list files"]


Single Thread:

main = do
print "start"
exitFailure
print "nothing to show"

That code will set exit code to 0
e.g. echo $? => 0 main = do print "start" exitWith(ExitFailure 3) print "nothing to show" e.g. echo$? => 3
You can use exitWith(...) to set exit code whatever you want

import Control.Exception.Safe
import System.Exit
import Control.Concurrent

main = do
print "start"
forkIO(exitWith(ExitFailure 3))
print "nothing to show"

echo $? => 0 You do not get 3 now, plz take a look at exitWith::ExitCode -> IO a forkIO creates worker thread, and the exit code does not propagate to the parent thread main = do print "start" concurrently (exitWith(ExitFailure 3))$ do
print "nothing to show"

echo $? => 0 The exit code propagates to parent thread with concurrently,  zip and unzip in Haskell zip ["dog", "cat"] [1, 2] => [("dog", 1"), ("cat", 2)] unzip [("dog", 1"), ("cat", 2)] => ["dog", "cat"] [1, 2]  Simple Typeclass in Haskell 1. define a typeclass Shape with two constructors Shape a is type class or type constructor Circle Float Float Float is one constructor with three floats or value constructor Rectangle Float Float Float Float is value constructor with four floats  -- Typeclass in Haskell -- Shape is similar to interface in Java(but it is different) -- Circle and Rectangle are similar to class in Java -- sum type data Shape = Circle Float Float Float | Rectangle Float Float Float Float deriving (Show) area::Shape->Float area(Circle _ _ r) = pi*r^2 area(Rectangle topx topy bottomx bottomy) = abs(bottomy - topy)*abs(bottomx - topx) -- or data Shape a = Circle a a a | Rectangle a a a a deriving (Show) area::Shape Float -> Float area (Circle _ _ r) = pi*r^2 area (Rectangle topx topy bottomx bottomy) = abs(bottomy - topy)*abs(bottomx - topx) main = do print("Typeclass in Haskell") print(area$ Circle 3 4 5)
print(area $Rectangle 3 4 5 8)  Equation Guarding in Haskell --Guards in Haskell myGuards x | x < 0 = -1 | x > 0 = 1 | otherwise = 0 main = do print (myGuards(3)) -- multiple logical conditions guard in Haskell myGuards x | x > 0 && x < -10 = -1 | x > 0 && x < -2 = 1 | otherwise = 0 main = do print (myGuards(-3))  Sum all integers in a list in Haskell -- Sum all the integer in a list sumlist::[Int]->Int sumlist [] = 0 sumlist (x:xs) = x+sumlist(xs)  Palindrome in Haskell isPalindrome:: Eq a=>[a]->Bool isPalindrome [a] = True isPalindrome [x,y] = x == y isPalindrome (x:xs) = x == (last xs) && isPalindrome(init(xs)) main = do print ("aba="++ show(isPalindrome("aaa")))  Maximum of a list --find the maximum of a list maximum1'::(Ord a) =>[a]->a maximum1' [] = error "empty list" maximum1' [x] = x maximum1' (x:xs) | x > maxTail = x | otherwise = maxTail where maxTail = maximum1' xs --find the maximum of a list without where maximum2'::(Ord a) =>[a]->a maximum2' [] = error "empty list" maximum2' [x] = x maximum2' (x:xs) = max x (maximum2' xs)  If else in Haskell --if else in Haskell mySign x = if x < 0 then -1 else if x > 0 then 1 else 0 main = do print (mySign(3))  let binding in Haskell -- let binding in Haskell -- ax^2 + bx + c = 0 -- Find the root of the quadratic equation roots a b c = let discriminane= sqrt(b^2 - 4*a*c) in ( (-b + discriminant)/(2*a), (-b - discriminant)/(2*a)) main = do print (roots 3 9 1)  Print all permutation in Haskell myperm:: Eq a =>[a]->[[a]] myperm [] = [[]] myperm xs = [ x:ys | x <- xs , ys <- myperm(delete x xs)]  Tree Traversal: Preorder Inorder Postorder/DFS in Haskell  There are preorder inorder and postorder traversals preorder: + a b inorder: a + b postorder: a b + There are many applications from these traversals 1. We can rebuild the tree with preorder and inorder 2. inorder is the sorted order in Binary Search Tree 3. Quick sort is related to preorder 4. inorder can be used to check whether a BT is BST 5. postorder can be implemented with two stacks with recursion 6. Merge sort is essential the postorder 7. Quick sort is essential the preorder 8. preorder and postorder can be used in serialization and deseriazation  data Tree a = Empty | Node a (Tree a) (Tree a) deriving Show mytree = Node 'R' (Node 'S' (Node 'T' Empty Empty) (Node 'U' Empty Empty) ) (Node 'T' (Node 'V' Empty Empty) (Node 'W' Empty Empty) ) preorderTraversal::Tree a ->[a] preorderTraversal Empty = [] preorderTraversal (Node a l r) = a:(preorderTraversal l) ++ (preorderTraversal r) inorderTraversal::Tree a ->[a] inorderTraversal Empty = [] inorderTraversal (Node a l r) = (inorderTraversal l) ++[a]++(inorderTraversal r) postorderTraversal::Tree a ->[a] postorderTraversal Empty = [] postorderTraversal(Node a l r) = (postorderTraversal l) ++ (postorderTraversal r) ++ [a]  More Typeclass 2 in Haskell --Typeclass 2 in Haskell data Point = Point Float Float deriving(Show) data Shape = Circle Point Float | Rectangle Point Point deriving (Show) area::Shape->Float area(Circle _ r) = pi*r^2 area(Rectangle (Point topx topy) (Point bottomx bottomy)) = abs(bottomy - topy)*abs(bottomx - topx) main = do print(area$ Circle (Point 1 4) 3)
print(area $Rectangle (Point 1 4) (Point 2 7))  More Typeclass 3 in Haskell -- Library/WebServer/Documents/zsurface/github/haskell/typeclass3.hs -- Typeclass in Haskell -- Shape is similar to interface in Java -- Circle and Rectangle are similar to class in Java data Point = Point Float Float deriving(Show) data Shape = Circle Float Float Float | Rectangle Float Float Float Float deriving (Show) --call constructor circleShape::Shape circleShape = Circle 1 2 3 rectangleShape = Rectangle 1 2 3 4 --list of typeclasses listShape::[Shape] listShape = [Circle 1 2 3, Rectangle 1 2 3 4] main = do print(circleShape) print(rectangleShape) print(listShape)  More Typeclass 4 in Haskell --Library/WebServer/Documents/zsurface/github/haskell/typeclass4.hs --Deconstructing type in Haskell data Point = Point Float Float deriving(Show) data Shape = Circle Float Float Float | Rectangle Float Float Float Float deriving (Show) showShape::Shape->String showShape(Circle x y radial) = show(x) ++ " "++show(y)++ " " ++ show(radial) showShape(Rectangle x1 y1 x2 y2 ) = show(x1) ++ " "++show(y1)++ " " ++ show(x2)++" "++show(y2) main = do print(showShape(Circle 3 4 5)) print(showShape(Rectangle 1 2 3 4))  More Typeclass 5 in Haskell data Foo = Bar | Baz Int myfun::Foo->Int myfun Bar = 1 myfun (Baz x) = x - 1  More Typeclass 6 in Haskell data Person = Person String Integer String deriving Show getAge::Person->Integer getAge(Person _ n _) = n getName::Person->String getName(Person name _ _) = name getInfo::Person->String getInfo p@(Person _ _ _) = "(" ++ show(p) ++ ")"  More Typeclass 7 in Haskell data Point a = Pt a a deriving(Show) dist::Point Int-> Point Int -> Int dist (Pt x0 y0) (Pt x1 y1) = (x0-x1)*(x0-x1) + (y0-y1)*(y0-y1) main = do print (Pt 1 3 :: Point Int) print$ dist(Pt 1 4) (Pt 3 4)

            data Point a = Pt a a
norm::(Point Int)->(Point Int)->Int
norm(Pt a1 b1) (Pt a2 b2)  = a1*a2 + b1*b2

dist::(Point Int)->(Point Int)->Int
dist (Pt a1 b1) (Pt a2 b2) = norm (Pt a1 b1) (Pt a2 b2)

// in Java
class Point<T>{
Point(T t1, T t2){}
}

Following is java interface and two classes
interface Shape
{
public double getArea();
}

class Circle implements Shape
{
double x;
double y;
public Circle(double x, double y, double radial)
{
this.x = x;
this.y = y;
}
public double getArea()
{
}
}

class Rectangle implements Shape
{
double topx;
double topy;
double bottomx;
double bottomy;
public Rectangle(double topx, double topy, double bottomx, double bottomy)
{
this.topx = topx;
this.topy = topy;
this.bottomx= bottomx;
this.bottomy= bottomy;
}
public double getArea()
{
return Math.abs(bottomx - topx)*Math.abs(bottomy - topy);
}
}

f::()->()
f _ = ()

f1::Bool->Bool
f1 _ = () == ()

f2::Bool->Bool
f2 x = (1 == 2) == True

f3::Bool->Bool
f3 _ = False

f4::Bool->Bool
f4 undefined = False

f5::Bool->Bool
f5 undefined = undefined

f6::Bool->Bool
f6 _ = undefined

f7::Bool->Bool
f7 = undefined

f8::Bool->Bool
f8 = \x -> undefined

f9::Int->Int
f9 = \x -> x + 1

f10::Int->Int
f10 = \x -> f9 x

f11::Int->Int
f11 x = x + 1 + (f9 1)

f12::Int->Int
f12 = \x -> (x + 1)

-- function composition foldable
f1::Int->Int
f1 a = a * 10

f2::Int->Int
f2 a = a - 10
compose = (f1.f2) 10

-- identity = []
concat2::[[Int]]->[Int]
concat2 vals = foldr (\val acc->val ++ acc) [] vals
-- concat2 [[1, 2], [3, 4]] =  [1, 2] ++ ([3, 4] ++ []) = [1, 2, 3, 4]

fo = foldr ((:)) [] [1, 2, 3]
-- output [1, 2, 3]

-- alternative definition of map
map'::(a->b)->[a]->[b]
map' f = foldr ((:).f) []
-- map' (1+) 1 [1, 2, 3] = [2, 3, 4]

-- identity = 0
sum1::[Int]->Int
sum1 vals = foldl (\x y-> x + y) 0 vals

-- identity = 1
product1::[Int]->Int
product1 vals = foldl (\x y-> x * y) 1 vals

-- identity = []
concat1::[[Char]]->String
concat1 vals = foldl (\x y->concat[x, y]) [] vals

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

-- instance class
data Myfoo = Myfoo{int::Integer, str::String}
instance Eq Myfoo where
(Myfoo int1 str1) == (Myfoo int2 str2) = (int1 == int2) && str1 == str2

data MyMaybe a = MyJust a | MyNothing deriving (Show)

instance Functor MyMaybe where
fmap f MyNothing = MyNothing
fmap f (MyJust a) = MyJust (f a)

            Algebraic Data type - Algebraic refers to the property that an algebraic data type is created
by algebraic operations such as sum and product. There
Sum is alternation[logic OR operator][A | B, meaning A or B but not both, e.g. enum in Java/C++]
Product is combination[logic AND operator][A B, meaning A and B together, e.g. tuple, record]

data Maybe a = Just a | Nothing
There are two constructors:
Just is one constructor with one parameter a
Nothing is other constrcutor with no parameter

-- Try to simulate Maybe in Java
interface Maybe{ }
class Just implements Maybe{
public Just(String s){ }
}
class Nothing implements Maybe{
public Nothing(){}
}

-- City is similar to Enum in Java/C++
-- public enum City
-- {
--  PaloAlto,
--  MountainView
--  Sunnyvale
-- }
-- City city = MountainView
--
data City = PaloAlto
| MountainView
| Sunnyvale
deriving(Show)

city::City
city = MountainView

listofCity::[City]
listofCity = [PaloAlto, Sunnyvale, MountainView]

isGoodCity::City -> Bool
isGoodCity PaloAlto     = True
isGoodCity MountainView = True
isGoodCity Sunnyvale    = False

data MyType = Failture
| OK (Double)
deriving Show

fa = Failture
ok = OK 3.31

safeDiv::Double->Double->MyType
safeDiv _ 0 = Failture
safeDiv x y = OK(x/y)

f::String->MyType
f   ""    = Failture
f   "cat" = OK (100)
f   "dog" = OK (200)
f   _     = OK 3.3

main = do
print (isGoodCity PaloAlto)
print (isGoodCity Sunnyvale)
print (fa, ok)
print (f "")
print (f "cat")
print (f "dog")
print (safeDiv 3 0)
print (safeDiv 3 4)


-- folding examples
main = do
print(foldr (-) 1 [2, 3, 4]) -- output 2
print(foldl (-) 1 [2, 3, 4]) -- output -8

let fr = foldr (\x y->concat["(",x,"-",y,")"]) "1" (map show [2, 3, 4])
print(fr)
--output "(2-(3-(4-1)))"

let fl = foldl (\x y->concat["(",x,"-",y,")"]) "1" (map show [2, 3, 4])
print(fl)
--output "(((1-2)-3)-4)"


// foldr (-) 1 [2, 3, 4]  output = 2
public static int foldr(int n){
if(n == 4){
return n - 1;
}
else{
return n - foldr(n + 1);
}
}
// foldl (-) 1 [2, 3, 4] output = -8
public static int foldl(int n){
if(n == 2){
return 1 - n;
}else{
return foldl(n-1) - n;
}
}


Map can be defined by foldr, let's look at the definition of map in Haskell
            map::(a->b)->[a]->[b]
map f (x:xs) = f x : map f xs

foldr f a [n1, n2, n3] = n1 f (n2 f (a f n3))

-- e.g.
foldr (+) 1 [2, 3, 4] = 2 + (3 + (4 + 1))

foldr ((:)) [] [2, 3, 4] = 2:3:4:[] = [2, 3, 4]

map'::(a->b)->[a]->[b]
map' f = foldr((:)) []

-- remove brackets
foldr (++) [] [["a", "b"], ["c"]]
-- output ["a", "b", "c"]


let expressions and where causes local variable binding in Haskell
-- where cause and let binding in Haskell
--
-- ax^2 + bx + c = 0
-- Find the root of the quadratic equation

roots a b c =
let discriminant= sqrt(b^2 - 4*a*c)
in ( (-b + discriminant)/(2*a), (-b - discriminant)/(2*a))

longest []     = []
longest (x:xs) = if length x > length (longest xs) then x else (longest xs)

newlongest []     = []
newlongest (x:xs) = if length x > length max then x else max
where max = newlongest xs

main =  do
print (roots 3 9 1)
print (longest [[],[1],[1,2]])
print (newlongest [[],[1],[1,2]])

Case in Haskell is like switch statement in Java, except it is more beautiful!
suffixed n = show n ++ suffix
where
suffix
| n mod 100 div 10 == 1 = "th"
| otherwise = case n mod 10 of
1 -> "st"
2 -> "nd"
3 -> "rd"
_ -> "th"

result = map suffixed [1..40]

fun x = case x mod 2 of
0 -> "even"
_ -> "odd"

| x < 10 = "less then 10"
| x == 10 = "equal 10"
| otherwise = case x mod 2 of
0 -> "greater than 10 even"
_ -> "greater than 10 and odd"

First OpenGL Hello World in Haskell
import Graphics.UI.GLUT

myPoints::[(GLfloat, GLfloat, GLfloat)]
myPoints = [(sin(2*pi*k/12), cos(2*pi*k/12), 0) | k <- [1..12]]

main::IO()
main = do
(_progName, _args) <- getArgsAndInitialize
_window <- createWindow "Hello World"
displayCallback $= display mainLoop display::DisplayCallback display = do clear [ ColorBuffer] renderPrimitive Triangles$ mapM_ (\(x, y, z) -> vertex $Vertex3 x y z) myPoints flush  OpenGL Vertex and Color in Haskell import Graphics.UI.GLUT main::IO() main = do (_progName, _args) <- getArgsAndInitialize _window <- createWindow "OpenGL Vertex and Color in Haskell" displayCallback$= display
mainLoop

display::DisplayCallback
display = do
let color3f r g b = color $Color3 r g (b::GLfloat) vertex3f x y z = vertex$ Vertex3 x y (z::GLfloat)
clear [ColorBuffer]
renderPrimitive Quads $do color3f 1 1 0 vertex3f 0 0 0 vertex3f 0 0.2 0 vertex3f 0.2 0.2 0 vertex3f 0.2 0 0 color3f 0 1 0.5 vertex3f 0 0 0 vertex3f 0 (-0.2) 0 vertex3f 0.2 (-0.2) 0 vertex3f 0.2 0 0 color3f 0 0.5 1 vertex3f 0 0 0 vertex3f 0 (-0.2) 0 vertex3f (-0.2) (-0.2) 0 vertex3f (-0.2) 0 0 color3f 0.3 0.3 0.3 vertex3f 0 0 0 vertex3f 0 0.2 0 vertex3f (-0.2) 0.2 0 vertex3f (-0.2) 0 0 flush  Map Examples in Haskell Haskell Map Example map::(a->b) -> [a] -> [b] map f xs is the list obtained by apply f to each element of xs map f [1..3] = [f(1) f(2) f(3)] map f [x1, x2, ...] = [f(x1), f(x2), ...] f::Int->Int->Int f a b = a - b main = do print(map (+ 2) [1..3]) -- (2 + 1) (2 + 2) (2 + 3) print(map (2 -) [1..3]) -- (2 - 1) (2 - 2) (2 - 2) print(map ((-) 2) [1..3]) -- (2 - 1) (2 - 2) (2 - 3) print(map (mod 2) [1..3]) -- (2 mod 1) (2 mod 2) (2 mod 3) print(map (mod 2) [1..3]) -- (1 mod 2) (2 mod 2) (3 mod 2) print(map (/ 2) [1..3]) -- (2 / 1) (2 / 2) (2 / 3) print(map (* 2) [1..3]) -- (2 * 1) (2 * 2) (2 * 3) print(map (f 10) [1..3]) -- (10 - 1) (10 - 2) (10 - 3) print(map (f 10) [1..3]) -- (1 - 10) (2 - 10) (3 - 10)  Unwrap Monad, Find Index and SubString in Haskell 1. Unwrap the Maybe Monad 2. Find the Index of Char in a String 3. SubString of two Indexes of string Similar Haskell functions "abc" !! 1 -> 'b' elemIndex 'a' "ba" -> Just 1 isInfixOf "ab" "efab" -> True findIndex (== 'a') "bac" -> 1 findIndex (> 'a') "bac" -> 2 --unwrap Maybe open::Maybe a->a open Nothing = error "error" open (Just a) = a -- find the index of Char findChar::Char->String->Int findChar _ [] = -1 findChar a (x:xs) = if a /= x then ((findChar a xs) + 1) else 0 -- substring from l to r range::Int->Int->String->String range l r str = if l >= length str || r >= length str then [] else (if l <= r then str !! l :[] ++ range (l+1) r str else [])  Regular Expression in Haskell mymatch regex f s = case matchRegexAll regex s of Nothing->s Just(before, matched, after, l)->before ++ f l ++ after t1 = filter (=~ "is") ["this is it"] -- ["this is it"] ignoreCase = makeRegexOpts(defaultCompOpt .|. compIgnoreCase) defaultExecOpt t3 = filter (matchTest$ ignoreCase "apple") ["My Apple"]
-- ["My Apple"]
t4 = subRegex(mkRegex "p+") "apple sucks" "PP"
-- aPPle sucks
t5 = mymatch(mkRegex "man([0-9])") (\[n]->"woman" ++ n) "/cat/man3/dog.txt"
-- "/cat/woman3/dog.txt"