PROGRAMMING IN HASKELL

1. PROGRAMMING IN HASKELL Chapter 9 - Higher-Order Functions, Functional Parsers

2. Introduction A function is called higher-order if it takes a function as an argument or returns a function as a result. twice :: (a  a)  a  a twice f x = f (f x) twice is higher-order because it takes a function as its first argument.

3. Why Are They Useful? • Common programming idioms can be encoded as functions within the language itself. • Domain specific languages can be defined as collections of higher-order functions. • Algebraic properties of higher-order functions can be used to reason about programs.

4. The Map Function The higher-order library function called map applies a function to every element of a list. map :: (a  b)  [a]  [b] For example: > map (+1) [1,3,5,7] [2,4,6,8]

5. The map function can be defined in a particularly simple manner using a list comprehension: map f xs = [f x | x  xs] Alternatively, for the purposes of proofs, the map function can also be defined using recursion: map f [] = [] map f (x:xs) = f x : map f xs

6. The Filter Function The higher-order library function filter selects every element from a list that satisfies a predicate. filter :: (a  Bool)  [a]  [a] For example: > filter even [1..10] [2,4,6,8,10]

7. Filter can be defined using a list comprehension: filter p xs = [x | x  xs, p x] Alternatively, it can be defined using recursion: filter p [] = [] filter p (x:xs) | p x = x : filter p xs | otherwise = filter p xs

8. The Foldr Function A number of functions on lists can be defined using the following simple pattern of recursion: f [] = v f (x:xs) = x  f xs f maps the empty list to some value v, and any non-empty list to some function  applied to its head and f of its tail.

9. For example: v = 0  = + sum [] = 0 sum (x:xs) = x + sum xs v = 1  = * product [] = 1 product (x:xs) = x * product xs v = True  = && and [] = True and (x:xs) = x && and xs

10. The higher-order library function foldr (fold right) encapsulates this simple pattern of recursion, with the function  and the value v as arguments. For example: sum = foldr (+) 0 product = foldr (*) 1 or = foldr (||) False and = foldr (&&) True

11. Foldr itself can be defined using recursion: foldr :: (a  b  b)  b  [a]  b foldr f v [] = v foldr f v (x:xs) = f x (foldr f v xs) However, it is best to think of foldr non-recursively, as simultaneously replacing each (:) in a list by a given function, and [] by a given value.

12. = foldr (+) 0 [1,2,3] = foldr (+) 0 (1:(2:(3:[]))) = 1+(2+(3+0)) = 6 For example: sum [1,2,3] Replace each (:) by (+) and [] by 0.

13. = foldr (*) 1 [1,2,3] = foldr (*) 1 (1:(2:(3:[]))) = 1*(2*(3*1)) = 6 For example: product [1,2,3] Replace each (:) by (*) and [] by 1.

14. Other Foldr Examples Even though foldr encapsulates a simple pattern of recursion, it can be used to define many more functions than might first be expected. Recall the length function: length :: [a]  Int length [] = 0 length (_:xs) = 1 + length xs

15. = length (1:(2:(3:[]))) = 1+(1+(1+0)) = Hence, we have: 3 length = foldr (_ n  1+n) 0 For example: length [1,2,3] Replace each (:) by _ n  1+n and [] by 0.

16. = reverse (1:(2:(3:[]))) = (([] ++ [3]) ++ [2]) ++ [1] = [3,2,1] Now recall the reverse function: reverse [] = [] reverse (x:xs) = reverse xs ++ [x] For example: Replace each (:) by x xs  xs ++ [x] and [] by []. reverse [1,2,3]

17. Hence, we have: reverse = foldr (x xs  xs ++ [x]) [] Finally, we note that the append function (++) has a particularly compact definition using foldr: Replace each (:) by (:) and [] by ys. (++ ys) = foldr (:) ys

18. Why Is Foldr Useful? • Some recursive functions on lists, such as sum, are simpler to define using foldr. • Properties of functions defined using foldr can be proved using algebraic properties of foldr, such as fusion and the banana split rule. • Advanced program optimisations can be simpler if foldr is used in place of explicit recursion.

19. Other Library Functions The library function (.) returns the composition of two functions as a single function. (.) :: (b  c)  (a  b)  (a  c) f . g = x  f (g x) For example: odd :: Int  Bool odd = not . even

20. The library function all decides if every element of a list satisfies a given predicate. all :: (a  Bool)  [a]  Bool all p xs = and [p x | x  xs] For example: > all even [2,4,6,8,10] True

21. Dually, the library function any decides if at least one element of a list satisfies a predicate. any :: (a  Bool)  [a]  Bool any p xs = or [p x | x  xs] For example: > any isSpace "abc def" True

22. The library function takeWhile selects elements from a list while a predicate holds of all the elements. takeWhile :: (a  Bool)  [a]  [a] takeWhile p [] = [] takeWhile p (x:xs) | p x = x : takeWhile p xs | otherwise = [] For example: > takeWhile isAlpha "abc def" "abc"

23. Dually, the function dropWhile removes elements while a predicate holds of all the elements. dropWhile :: (a  Bool)  [a]  [a] dropWhile p [] = [] dropWhile p (x:xs) | p x = dropWhile p xs | otherwise = x:xs For example: > dropWhile isSpace " abc" "abc"

24. (1) What are higher-order functions that return functions as results better known as? (2) Express the comprehension [f x | x  xs, p x] using the functions map and filter. (3) Redefine map f and filter p using foldr. Exercises

25. + means  4 2 2 3 What is a Parser? A parser is a program that analyses a piece of text to determine its syntactic structure. 23+4

26. Haskell programs Shell scripts HTML documents Hugs Unix Explorer parses Where Are They Used? Almost every real life program uses some form of parser to pre-process its input.

27. The Parser Type In a functional language such as Haskell, parsers can naturally be viewed as functions. type Parser = String  Tree A parser is a function that takes a string and returns some form of tree.

28. However, a parser might not require all of its input string, so we also return any unused input: type Parser = String  (Tree,String) A string might be parsable in many ways, including none, so we generalize to a list of results: type Parser = String  [(Tree,String)]

29. Finally, a parser might not always produce a tree, so we generalize to a value of any type: type Parser a = String  [(a,String)] Note: • For simplicity, we will only consider parsers that either fail and return the empty list of results, or succeed and return a singleton list.

30. Basic Parsers • The parser item fails if the input is empty, and consumes the first character otherwise: item :: Parser Char item = inp  case inp of []  [] (x:xs)  [(x,xs)]

31. The parser failure always fails: failure :: Parser a failure = inp  [] • The parser return v always succeeds, returning the value v without consuming any input: return :: a  Parser a return v = inp  [(v,inp)]

32. The parser p +++ q behaves as the parser p if it succeeds, and as the parser q otherwise: (+++) :: Parser a  Parser a  Parser a p +++ q = inp  case p inp of []  parse q inp [(v,out)]  [(v,out)] • The function parse applies a parser to a string: parse :: Parser a  String  [(a,String)] parse p inp = p inp

33. Examples The behavior of the five parsing primitives can be illustrated with some simple examples: % hugs Parsing > parse item "" [] > parse item "abc" [('a',"bc")]

34. > parse failure "abc" [] > parse (return 1) "abc" [(1,"abc")] > parse (item +++ return 'd') "abc" [('a',"bc")] > parse (failure +++ return 'd') "abc" [('d',"abc")]

35. Note: • The library file Parsing is available on the web from the Programming in Haskell home page. • For technical reasons, the first failure example actually gives an error concerning types, but this does not occur in non-trivial examples. • The Parser type is a monad, a mathematical structure that has proved useful for modeling many different kinds of computations.

36. Sequencing A sequence of parsers can be combined as a single composite parser using the keyword do. For example: p :: Parser (Char,Char) p = do x  item item y  item return (x,y)

37. Note: • Each parser must begin in precisely the same column. That is, the layout rule applies. • The values returned by intermediate parsers are discarded by default, but if required can be named using the  operator. • The value returned by the last parser is the value returned by the sequence as a whole.

38. If any parser in a sequence of parsers fails, then the sequence as a whole fails. For example: > parse p "abcdef" [((’a’,’c’),"def")] > parse p "ab" [] • The do notation is not specific to the Parser type, but can be used with any monadic type.

39. Derived Primitives • Parsing a character that satisfies a predicate: sat :: (Char  Bool)  Parser Char sat p = do x  item if p x then return x else failure

40. Parsing a digit and specific characters: digit :: Parser Char digit = sat isDigit char :: Char  Parser Char char x = sat (x ==) • Applying a parser zero or more times: many :: Parser a  Parser [a] many p = many1 p +++ return []

41. Applying a parser one or more times: many1 :: Parser a -> Parser [a] many1 p = do v  p vs  many p return (v:vs) • Parsing a specific string of characters: string :: String  Parser String string [] = return [] string (x:xs) = do char x string xs return (x:xs)

42. Example We can now define a parser that consumes a list of one or more digits from a string: p :: Parser String p = do char '[' d  digit ds  many (do char ',' digit) char ']' return (d:ds)

43. For example: > parse p "[1,2,3,4]" [("1234","")] > parse p "[1,2,3,4" [] Note: • More sophisticated parsing libraries can indicate and/or recover from errors in the input string.

44. Arithmetic Expressions Consider a simple form of expressions built up from single digits using the operations of addition + and multiplication *, together with parentheses. We also assume that: • * and + associate to the right; • * has higher priority than +.

45. Formally, the syntax of such expressions is defined by the following context free grammar: expr term '+' expr term term factor '*' term  factor factordigit  '(' expr ')‘ digit '0'  '1'  '9'

46. However, for reasons of efficiency, it is important to factorise the rules for expr and term: expr term ('+' expr) term factor ('*' term) Note: • The symbol  denotes the empty string.

47. It is now easy to translate the grammar into a parser that evaluates expressions, by simply rewriting the grammar rules using the parsing primitives. That is, we have: expr :: Parser Int expr = do t  term do char '+' e  expr return (t + e) +++ return t

48. term :: Parser Int term = do f  factor do char '*' t  term return (f * t) +++ return f factor :: Parser Int factor = do d  digit return (digitToInt d) +++ do char '(' e  expr char ')' return e

49. Finally, if we define eval :: String  Int eval xs = fst (head (parse expr xs)) then we try out some examples: > eval "2*3+4" 10 > eval "2*(3+4)" 14

50. (2) Extend the expression parser to allow the use of subtraction and division, based upon the following extensions to the grammar: (1) Why does factorising the expression grammar make the resulting parser more efficient? Exercises expr term ('+' expr '-' expr) term factor ('*' term '/' term)