[英]Haskell/Trifecta: Parsing completely optional semicolons without polluting AST
[英]AST and Parsing in Haskell
我有一個任務,我無法弄清楚如何定義答案。
寫函數exp:: [String] -> (AST, [String])
AST
:
x
是數字,則應該表示Number x
。 Atom x
。 [AST]
。 這樣輸出將是:
exp (token "(hi (4) 32)")
> (List [Atom "hi", List [Number 4], Number 32], [])
exp (token "(+ 3 42 654 2)")
> (List [Atom "+", Number 3, Number 42, Number 654, Number 2], [])
exp (token "(+ 21 444) junk")
> (List [Atom "+", Number 21, Number 444], ["junk"])
我已經有了令牌函數, token :: String -> [String]
,它就是一個列表。
exp :: [String] -> (AST, [String])
exp [] = error "Empty list"
exp (x:xs) | x == ")" = error ""
| x == "(" = let (e, ss') = exp xs in (List [getAst xs], ss')
| x == "+" = let (e, ss') = exp xs in (Atom (read x), ss')
| x == "-" = let (e, ss') = exp xs in (Atom (read x), ss')
| otherwise = exp xs`
exp
函數如下所示:
getAst :: [String] -> AST
getAst [] = error ""
getAst (x:xs)
| x == ")" = error ""
| x == "(" = (List [getAst xs])
| isAtom x = (Atom x)
| isNum x = (Number (read x))
| otherwise = getAst xs`
getAst
函數的位置:
getAst :: [String] -> AST getAst [] = error "" getAst (x:xs) | x == ")" = error "" | x == "(" = (List [getAst xs]) | isAtom x = (Atom x) | isNum x = (Number (read x)) | otherwise = getAst xs`
(是的,我是Haskell的初學者......)
我想我可以試着幫助你一點。
表示問題的方式你應該能夠通過查看下一個輸入/令牌並從那里決定去哪里來做到這一點。
數據表示為[String] -> (Ast, [String])
我假設它是一個常見的解析器,解析器嘗試讀取輸入的某些部分並將解析/轉換的輸出與其余部分一起返回輸入它沒有變換(所以只是元組的兩個解析 - Ast
和輸入的其余部分)。
因為你沒有包含它,我認為它是:
data Ast
= Number Int
| Atom String
| List [Ast]
deriving Show
我需要一些東西:
import Prelude hiding (exp)
import Control.Applicative ((<$>))
import Data.Maybe (fromJust, isJust)
我必須隱藏exp
因為我們想將它用作函數名。
然后我想要一個Maybe
fmap
,所以我包含來自Control.Applicative
的運算符。 這真的就是這個,如果您之前沒有看到它:
f <$> Nothing = Nothing
f <$> Just a = Just (f a)
我想為Maybe
一些幫手:
isJust
要檢查是否Just _
fromJust
為了得到a
來自Just a
最后我需要這個幫助函數來read
更安全的東西:
tryRead :: (Read a) => String -> Maybe a
tryRead input =
case readsPrec 0 input of
(a,_):_ -> Just a
_ -> Nothing
這將嘗試在這里讀到一個數字- returing Just n
,如果n是一個數字, Nothing
其他。
這是一個未完成的第一個問題:
exp :: [String] -> (Ast, [String])
exp (lookat:rest)
| isJust number = (fromJust number, rest)
| lookat == "(" = parseList rest []
where number = Number <$> tryRead lookat
parseList :: [String] -> [Ast] -> (Ast, [String])
parseList inp@(lookat:rest) acc
| lookat == ")" = (List (reverse acc), rest)
| otherwise = let (el, rest') = exp inp
in parseList rest' (el:acc)
正如你所看到的那樣,我只是根據lookat
分支,但略有不同:
如果我看到一個數字,我會返回數字和rest-token-list。 如果我看到一個(
我啟動另一個解析器parseList
。
parseList
將執行相同的操作: - 它查看第一個標記 - 如果標記是a )
它完成當前列表(它使用累加器技術)並返回。 - 如果不是,它使用現有的exp
解析器遞歸獲取列表的元素。
這是一個示例運行:
λ> let input = ["(", "2", "(", "3", "4", ")", "5", ")"]
λ> exp input
(List [Number 2,List [Number 3,Number 4],Number 5],[])
還有一些邊界情況你必須決定(如果沒有輸入令牌怎么辦?)。
當然,你必須為Atom
添加案例 - 完成這個例外。
好的 - 3個小時后,OP沒有再次辦理登機手續,所以我想我可以發布一個完整的解決方案。 我希望我沒有忘記任何邊緣情況,這肯定不是最有效的實現( tokens
浮現在腦海中) - 但OP給出了所有匹配的示例:
module Ast where
import Prelude hiding (exp)
import Control.Applicative ((<$>))
import Data.Char (isSpace, isControl)
import Data.Maybe (fromJust, isJust)
data Ast
= Number Int
| Atom String
| List [Ast]
| Empty
deriving Show
type Token = String
main :: IO ()
main = do
print $ parse "(hi (4) 32)"
print $ parse "(+ 3 42 654 2)"
print $ parseAst . tokens $ "(+ 21 444) junk"
parse :: String -> Ast
parse = fst . parseAst . tokens
parseAst :: [Token] -> (Ast, [Token])
parseAst [] = (Empty, [])
parseAst (lookat:rest)
| isJust number = (fromJust number, rest)
| lookat == "(" = parseList rest []
| otherwise = (Atom lookat, rest)
where number = Number <$> tryRead lookat
parseList :: [Token] -> [Ast] -> (Ast, [Token])
parseList [] _ = error "Syntax error: `)` not found"
parseList inp@(lookat:rest) acc
| lookat == ")" = (List (reverse acc), rest)
| otherwise = let (el, rest') = parseAst inp
in parseList rest' (el:acc)
tokens :: String -> [Token]
tokens = split ""
where split tok "" = add tok []
split tok (c:cs)
| c == '(' || c == ')' = add tok $ [c] : split "" cs
| isSpace c || isControl c = add tok $ split "" cs
| otherwise = split (tok ++ [c]) cs
add "" tks = tks
add t tks = t : tks
tryRead :: (Read a) => Token -> Maybe a
tryRead input =
case readsPrec 0 input of
(a,_):_ -> Just a
_ -> Nothing
λ> :main
List [Atom "hi",List [Number 4],Number 32]
List [Atom "+",Number 3,Number 42,Number 654,Number 2]
(List [Atom "+",Number 21,Number 444],["junk"])
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