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

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  • *PROGRAMMING IN HASKELLChapter 9 - Higher-Order Functions,Functional Parsers

  • 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 atwice f x = f (f x)twice is higher-order because ittakes a function as its first argument.

  • 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.

  • 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]


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

  • 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]


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

  • The Foldr Function*A number of functions on lists can be defined using the following simple pattern of recursion:f [] = vf (x:xs) = x f xsf 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.

  • *For example:sum [] = 0sum (x:xs) = x + sum xsand [] = Trueand (x:xs) = x && and xsproduct [] = 1product (x:xs) = x * product xsv = 0 = +v = 1 = *v = True = &&

  • *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

  • *Foldr itself can be defined using recursion:foldr :: (a b b) b [a] bfoldr f v [] = vfoldr 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.

  • *sum [1,2,3]For example:Replace each (:)by (+) and [] by 0.

  • *product [1,2,3]For example:Replace each (:)by (*) and [] by 1.

  • 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] Intlength [] = 0length (_:xs) = 1 + length xs

  • *length [1,2,3]Replace each (:) by _ n 1+n and [] by 0.For example:

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

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

  • 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.

  • 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 Boolodd = not . even

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


  • *Dually, the library function any decides if at leastone element of a list satisfies a predicate.any :: (a Bool) [a] Boolany p xs = or [p x | x xs]For example:> any isSpace "abc def"


  • *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"


  • *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:xsFor example:> dropWhile isSpace " abc"


  • Exercises*

  • What is a Parser?*A parser is a program that analyses a piece of text to determine its syntactic structure.23+4

  • Where Are They Used?*Almost every real life program uses some form of parser to pre-process its input.

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

  • *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)]

  • *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.

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

  • *The parser failure always fails:failure :: Parser afailure = inp []The parser return v always succeeds, returning the value v without consuming any input:return :: a Parser areturn v = inp [(v,inp)]

  • *The parser p +++ q behaves as the parser p if it succeeds, and as the parser q otherwise:(+++) :: Parser a Parser a Parser ap +++ 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

  • Examples*% hugs Parsing

    > parse item ""[]

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

  • *> parse failure "abc"[]

    > parse (return 1) "abc"[(1,"abc")]

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

    > parse (failure +++ return 'd') "abc"[('d',"abc")]

  • *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.

  • 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)

  • *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.

  • *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.

  • Derived Primitives*sat :: (Char Bool) Parser Charsat p = do x item if p x then return x else failureParsing a character that satisfies a predicate:

  • *digit :: Parser Chardigit = sat isDigit

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

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

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

  • *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.

  • 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 +.

  • *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'

  • *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.

  • *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 Intexpr = do t term do char '+' e expr return (t + e) +++ return t

  • *factor :: Parser Intfactor = do d digit return (digitToInt d) +++ do char '(' e expr char ')' return eterm :: Parser Intterm = do f factor do char '*' t term return (f * t) +++ return f

  • *Finally, if we defineeval :: String Inteval xs = fst (head (parse expr xs))then we try out some examples:> eval "2*3+4"10

    > eval "2*(3+4)"14

  • Exercises*expr term ('+' expr '-' expr )

    term factor ('*' term '/' term )