Re: [Haskell-cafe] Newbie: a parser for a list of objects?

-----Ursprüngliche Nachricht----- Von: "Dmitri O.Kondratiev" Gesendet: 26.03.07 16:44:12 An: haskell-cafe@haskell.org Betreff: [Haskell-cafe] Newbie: a parser for a list of objects?

Please see my questions inside comments {-- --} : Thanks!

--- module Parser where

import Data.Char

type Parse a b = [a] -> [(b, [a])]

{-- Newbie: a parser for a list of objects?

I am working with the section 17.5 "Case study: parsing expressions" of the book "Haskell The Craft of Functional Programming", where a parser for a list of objects is defined. I called this function pList in order to avoid confusion with 'list' as a term for data structure.

Please help me to understand how pList works (please, see the rest of the code at the end of this message): --}

pList :: Parse a b -> Parse a [b] pList p = (succeed []) `alt` ((p >*> pList p) `build` (uncurry (:)))

{-- First of all, I don't quite understand why there must be a choice ('alt') between the function ('succeed') that always returns an empty list and the other part? This results in adding [] to the front, why?

Well, if the parser p doesn't succeed, we don't want the whole thing to fail. And p will (almost certainly) fail when the end of input is reached. So without the alternative 'succeed []', we'd get pL1 dig "12" = [(('1':y),rem) | (y,rem) <- pL1 dig "2"] = [(('1':y),rem) | (y,rem) <- [(('2':z),rem2) | (z,rem2) <- pL1 dig ""]] = [(('1':y),rem) | (y,rem) <- [(('2':z),rem2) | (z,rem2) <- []] = [(('1':y),rem) | (y,rem) <- []] = [] because dig "" = []

I thought that 'simplified' version of pList should still work fine. Trying to prove this I wrote : --}

pL1 :: Parse a b -> Parse a [b] pL1 p = (p >*> pL1 p) `build` (uncurry (:))

{-- Which, as expected, does not work correctly - just gives an empty list [] - but I don't understand why:

because the parser eventually fails when the end of input is reached.

*Parser> t1 "12345" [] *Parser>

Also, I don't understand why the textbook version of pList gives this result:

*Parser> test "12345" [("","12345"),("1","2345"),("12","345"),("123","45"),("1234","5"),("12345","")]

That's because of the order of alt's arguments: (succeed [] `alt` p) inp = [([],inp)] ++ (p inp) with pList p = ((p >*> pList p) `build` (uncurry (:))) `alt` succeed [] the resulting list woulde be reversed.

*Parser>

In particular, I don't understand where the first element ("","12345") of the resulting list comes from?

I am trying to figure out how pList recursively unfolds. To my mind operators in the expression:

(succeed []) `alt`((p >*> pList p) `build` (uncurry (:)))

has the following execution order: 1) >*> 2) 'build' 3) 'alt'

No, the first argument of alt gets evaluated first, because (p1 `alt` p2) inp = (p1 inp) ++ (p2 inp), thus we need p1 inp first. Then we see we haven't hit bottom, so we need the second argument of (++) (resp. alt). So next we need to evaluate p, then pList p, combine the results of those with the second argument of build, uncurry (:).

It seems that operation >*> should be done as many times as many elements the input list has. Right?

Unfortunately not. Let's stay with pList dig. Say your input starts with n digits.

From the example above you can conjecture that length (pList dig inp) == (n+1). Now in the outermost (dig >*> pList dig) branch, you apply (pList dig) to an input beginning with (n-1) digits, returning a list of length n, to each element of this list you adjoin the first digit, resulting in n + (n-1) + ... + 1 = n*(n+1)/2 applications of (>*>). (Lesson: you need an exclusive choice, using the second parser only if the first one fails and a maximal munch combinator in your library, too)

Signature:

(>*>) :: Parse a b -> Parse a c -> Parse a (b, c)

implies that second argument of the expression:

p >*> pList p

should be of type 'Parse a c' but in this application it is of type 'Parse a b -> Parse a [b]'

c is [b], so p >*> pList p has type Parse a (b,[b]), then (p >*> pList p) `build` (uncurry (:)) has type Parse a [b]

How can that be? How recursion termination conditinon is expressed in pList?

recursion terminates when p fails. HTH, Daniel

--}

none :: Parse a b none inp = []

succeed :: b -> Parse a b succeed val inp = [(val, inp)]

suc:: b -> [a] -> [(b, [a])]

suc val inp = [(val, inp)]

spot :: (a -> Bool) -> Parse a a spot p [] = [] spot p (x:xs) | p x = [(x, xs)] | otherwise = []

alt :: Parse a b -> Parse a b -> Parse a b alt p1 p2 inp = p1 inp ++ p2 inp

bracket = spot (=='(') dash = spot (== '-') dig = spot isDigit alpha = spot isAlpha

infixr 5 >*>

(>*>) :: Parse a b -> Parse a c -> Parse a (b, c)

(>*>) p1 p2 inp = [((x,y), rem2) |(x, rem1) <- p1 inp, (y, rem2) <- p2 rem1]

build :: Parse a b -> (b -> c) -> Parse a c build p f inp = [ (f x, rem) | (x, rem) <- p inp]

test = pList dig t1 = pL1 dig

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