Write Yourself a Scheme in 48 Hours/Answers
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Chapter 1
editExercise 1
editmain :: IO () main = do args <- getArgs putStrLn ("Hello, " ++ args!!0 ++ " " ++ args!!1)
Exercise 2
editmain :: IO () main = do args <- getArgs print ((read $ args!!0) + (read $ args!!1))
The $
operator reduces the number of parentheses needed here. Alternatively you could write the function applications as read (args!!0)
.
Exercise 3
editmain :: IO () main = do putStrLn "What do they call thee at home?" name <- getLine putStrLn ("Ey up " ++ name)
Chapter 2
editSection 3 - Parsing
editExercise 1
editPart 1
editparseNumber :: Parser LispVal parseNumber = do x <- many1 digit (return . Number . read) x
Part 2
editIn order to answer this question, you need to do a bit of detective work! It is helpful to read up on do notation. Using the information there, we can mechanically transform the above answer into the following.
parseNumber = many1 digit >>= \x -> (return . Number . read) x
This can be cleaned up into the following:
parseNumber = many1 digit >>= return . Number . read
Exercise 2
editWe need to create a new parser action that accepts a backslash followed by either another backslash or a doublequote. This action needs to return only the second character.
escapedChars :: Parser Char escapedChars = do char '\\' -- a backslash x <- oneOf "\\\"" -- either backslash or doublequote return x -- return the escaped character
Once that is done, we need to make some changes to parseString.
parseString :: Parser LispVal parseString = do char '"' x <- many $ escapedChars <|> noneOf "\"\\" char '"' return $ String x
Exercise 3
editescapedChars :: Parser Char escapedChars = do char '\\' x <- oneOf "\\\"nrt" return $ case x of '\\' -> x '"' -> x 'n' -> '\n' 'r' -> '\r' 't' -> '\t'
Exercise 4
editFirst, it is necessary to change the definition of symbol.
symbol :: Parser Char symbol = oneOf "!$%&|*+-/:<=>?@^_~"
This means that it is no longer possible to begin an atom with the hash character. This necessitates a different way of parsing #t and #f.
parseBool :: Parser LispVal parseBool = do char '#' (char 't' >> return (Bool True)) <|> (char 'f' >> return (Bool False))
This in turn requires us to make changes to parseExpr.
parseExpr :: Parser LispVal parseExpr = parseAtom <|> parseString <|> parseNumber <|> parseBool
parseNumber need to be changed to the following.
parseNumber :: Parser LispVal parseNumber = parseDecimal1 <|> parseDecimal2 <|> parseHex <|> parseOct <|> parseBin
And the following new functions need to be added.
parseDecimal1 :: Parser LispVal parseDecimal1 = many1 digit >>= (return . Number . read)
parseDecimal2 :: Parser LispVal parseDecimal2 = do try $ string "#d" x <- many1 digit (return . Number . read) x
parseHex :: Parser LispVal parseHex = do try $ string "#x" x <- many1 hexDigit return $ Number (hex2dig x)
parseOct :: Parser LispVal parseOct = do try $ string "#o" x <- many1 octDigit return $ Number (oct2dig x)
parseBin :: Parser LispVal parseBin = do try $ string "#b" x <- many1 (oneOf "10") return $ Number (bin2dig x)
oct2dig x = fst $ readOct x !! 0 hex2dig x = fst $ readHex x !! 0 bin2dig = bin2dig' 0 bin2dig' digint "" = digint bin2dig' digint (x:xs) = let old = 2 * digint + (if x == '0' then 0 else 1) in bin2dig' old xs
Import the Numeric module to use the readOct and readHex functions.
Exercise 5
editdata LispVal = Atom String | List [LispVal] | DottedList [LispVal] LispVal | Number Integer | String String | Bool Bool | Character Char
parseCharacter :: Parser LispVal parseCharacter = do try $ string "#\\" value <- try (string "newline" <|> string "space") <|> do { x <- anyChar; notFollowedBy alphaNum ; return [x] } return $ Character $ case value of "space" -> ' ' "newline" -> '\n' otherwise -> (value !! 0)
The combination of anyChar and notFollowedBy ensure that only a single character is read.
Note that this does not actually conform to the standard; as it stands, "space" and "newline" must be entirely lowercase; the standard states that they should be case insensitive.
parseExpr :: Parser LispVal parseExpr = parseAtom <|> parseString <|> try parseNumber -- we need the 'try' because <|> try parseBool -- these can all start with the hash char <|> try parseCharacter
Exercise 6
editA possible solution for floating point numbers:
parseFloat :: Parser LispVal parseFloat = do x <- many1 digit char '.' y <- many1 digit return $ Float (fst.head$readFloat (x++"."++y))
Furthermore, add
try parseFloat
before parseNumber in parseExpr and the line
| Float Double
to the LispVal type.
Exercise 7
editRatio, using Haskell's Rational type:
parseRatio :: Parser LispVal parseRatio = do x <- many1 digit char '/' y <- many1 digit return $ Ratio ((read x) % (read y))
Additionally, import the Data.Ratio module, add
try parseRatio
before parseNumber in parseExpr and the line
| Ratio Rational
to the LispVal type.
Real is already implemented in the Float type from Exercise 6, unless I'm mistaken.
Complex using Haskell's Complex type:
toDouble :: LispVal -> Double toDouble(Float f) = realToFrac f toDouble(Number n) = fromIntegral n
parseComplex :: Parser LispVal parseComplex = do x <- (try parseFloat <|> parseDecimal) char '+' y <- (try parseFloat <|> parseDecimal) char 'i' return $ Complex (toDouble x :+ toDouble y)
As before, import the Data.Complex module, add
try parseComplex
before parseNumber and parseFloat in parseExpr and the line
| Complex (Complex Double)
to the LispVal type.
Section 4 - Recursive Parsers: Adding lists, dotted lists, and quoted datums
editExercise 1
editThese two are analogous to parseQuoted:
parseQuasiQuoted :: Parser LispVal parseQuasiQuoted = do char '`' x <- parseExpr return $ List [Atom "quasiquote", x]
parseUnQuote :: Parser LispVal parseUnQuote = do char ',' x <- parseExpr return $ List [Atom "unquote", x]
parseUnQuoteSplicing :: Parser LispVal parseUnQuoteSplicing = do char ',' char '@' x <- parseExpr return $ List [Atom "unquote-splicing", x]
Also add
<|> parseQuasiQuoted <|> parseUnQuote <|> parseUnQuoteSplicing
to parseExpr.
Exercise 2
editI chose to go with Arrays as described in Data.Array and used list-array conversions for array construction.
parseVector :: Parser LispVal parseVector = do arrayValues <- sepBy parseExpr spaces return $ Vector (listArray (0,(length arrayValues - 1)) arrayValues)
In order to use this, import Data.Array and add the following to the LispVal type:
| Vector (Array Int LispVal)
Add the following lines to parseExpr; before the parser for Lists and DottedLists.
<|> try (do string "#(" x <- parseVector char ')' return x)
Exercise 3
editThis took a fair amount of fiddling with sepBy
, endBy
and friends. I started by getting the (. degenerate)
dotted list to work and then went from there. This code tolerates trailing and leading spaces.
parseAnyList :: Parser LispVal parseAnyList = do P.char '(' optionalSpaces head <- P.sepEndBy parseExpr spaces tail <- (P.char '.' >> spaces >> parseExpr) <|> return (Nil ()) optionalSpaces P.char ')' return $ case tail of (Nil ()) -> List head otherwise -> DottedList head tail
Another implementation using more advanced functions from the Parsec library. spaces
is the one from this tutorial.
parseList :: Parser LispVal parseList = between beg end parseList1 where beg = (char '(' >> skipMany space) end = (skipMany space >> char ')')
parseList1 :: Parser LispVal parseList1 = do list <- sepEndBy parseExpr spaces maybeDatum <- optionMaybe (char '.' >> spaces >> parseExpr) return $ case maybeDatum of Nothing -> List list Just datum -> DottedList list datum
Alternative solution. spaces
is the spaces from Parsec and spaces1
is the spaces from this tutorial.
parseList :: Parser LispVal parseList = do char '(' >> spaces head <- parseExpr `sepEndBy` spaces1 do char '.' >> spaces1 tail <- parseExpr spaces >> char ')' return $ DottedList head tail <|> (spaces >> char ')' >> (return $ List head))
Chapter 3
editExercise 1
editHere is one way of adding a few of them.
primitives :: [(String , [LispVal] -> LispVal)] primitives = [("+" , numericBinop (+)) , ("-" , numericBinop (-)) , ("*" , numericBinop (*)) , ("/" , numericBinop div) , ("mod" , numericBinop mod) , ("quotient" , numericBinop quot) , ("remainder" , numericBinop rem) , ("symbol?" , unaryOp symbolp) , ("string?" , unaryOp stringp) , ("number?" , unaryOp numberp) , ("bool?", unaryOp boolp) , ("list?" , unaryOp listp)]
unaryOp :: (LispVal -> LispVal) -> [LispVal] -> LispVal unaryOp f [v] = f v
symbolp, numberp, stringp, boolp, listp :: LispVal -> LispVal symbolp (Atom _) = Bool True symbolp _ = Bool False numberp (Number _) = Bool True numberp _ = Bool False stringp (String _) = Bool True stringp _ = Bool False boolp (Bool _) = Bool True boolp _ = Bool False listp (List _) = Bool True listp (DottedList _ _) = Bool False listp _ = Bool False
Exercise 2
editunpackNum :: LispVal -> Integer unpackNum (Number n) = n unpackNum _ = 0
Exercise 3
editAdd symbol->string and string->symbol to the list of primitives, then:
symbol2string, string2symbol :: LispVal -> LispVal symbol2string (Atom s) = String s symbol2string _ = String "" string2symbol (String s) = Atom s string2symbol _ = Atom ""
This doesn't deal well with bad input, which is covered later.
Chapter 5
editExercise 1
editeval env (List [Atom "if", pred, conseq, alt]) = do result <- eval env pred case result of Bool False -> eval env alt Bool True -> eval env conseq _ -> throwError $ TypeMismatch "bool" pred
Exercise 2
editDefine a helper function that takes the equal/eqv function as an argument:
eqvList :: ([LispVal] -> ThrowsError LispVal) -> [LispVal] -> ThrowsError LispVal eqvList eqvFunc [(List arg1), (List arg2)] = return $ Bool $ (length arg1 == length arg2) && (all eqvPair $ zip arg1 arg2) where eqvPair (x1, x2) = case eqvFunc [x1, x2] of Left err -> False Right (Bool val) -> val
Now adjust the eqv clause:
eqv listPair@[List _, List _] = eqvList eqv listPair
And add clauses for List and DottedList to the equal function:
equal :: [LispVal] -> ThrowsError LispVal equal listPair@[List _, List _] = eqvList equal listPair equal [(DottedList xs x), (DottedList ys y)] = equal [List $ xs ++ [x], List $ ys ++ [y]] equal [arg1, arg2] = do primitiveEquals <- liftM or $ mapM (unpackEquals arg1 arg2) [AnyUnpacker unpackNum, AnyUnpacker unpackStr, AnyUnpacker unpackBool] eqvEquals <- eqv [arg1, arg2] return $ Bool $ (primitiveEquals || let (Bool x) = eqvEquals in x) equal badArgList = throwError $ NumArgs 2 badArgList
Exercise 3
editcond
editRoom for improvement here!
eval (List ((Atom "cond"):cs)) = do b <- (liftM (take 1 . dropWhile f) $ mapM condClause cs) >>= cdr car [b] >>= eval where condClause (List [p,b]) = do q <- eval p case q of Bool _ -> return $ List [q,b] _ -> throwError $ TypeMismatch "bool" q condClause v = throwError $ TypeMismatch "(pred body)" v f = \(List [p,b]) -> case p of (Bool False) -> True _ -> False
Another approach:
eval env (List (Atom "cond" : expr : rest)) = do eval' expr rest where eval' (List [cond, value]) (x : xs) = do result <- eval env cond case result of Bool False -> eval' x xs Bool True -> eval env value otherwise -> throwError $ TypeMismatch "boolean" cond eval' (List [Atom "else", value]) [] = do eval env value eval' (List [cond, value]) [] = do result <- eval env cond case result of Bool True -> eval env value otherwise -> throwError $ TypeMismatch "boolean" cond
Yet another approach, piggy-backing off of the already-implemented if:
eval form@(List (Atom "cond" : clauses)) = if null clauses then throwError $ BadSpecialForm "no true clause in cond expression: " form else case head clauses of List [Atom "else", expr] -> eval expr List [test, expr] -> eval $ List [Atom "if", test, expr, List (Atom "cond" : tail clauses)] _ -> throwError $ BadSpecialForm "ill-formed cond expression: " form
Yet another approach
eval (List ((Atom "cond") : alts)) = cond alts
cond :: [LispVal] -> ThrowsError LispVal cond ((List (Atom "else" : value : [])) : []) = eval value cond ((List (condition : value : [])) : alts) = do result <- eval condition boolResult :: Bool <- unpackBool result if boolResult then eval value else cond alts cond ((List a) : _) = throwError $ NumArgs 2 a cond (a : _) = throwError $ NumArgs 2 [a] cond _ = throwError $ Default "Not viable alternative in cond"
case
editThis solution requires LispVal to have a deriving (Eq) clause, in order to use the `elem` function.
eval form@(List (Atom "case" : key : clauses)) = if null clauses then throwError $ BadSpecialForm "no true clause in case expression: " form else case head clauses of List (Atom "else" : exprs) -> mapM eval exprs >>= return . last List ((List datums) : exprs) -> do result <- eval key equality <- mapM (\x -> eqv [result, x]) datums if Boolean True `elem` equality then mapM eval exprs >>= return . last else eval $ List (Atom "case" : key : tail clauses) _ -> throwError $ BadSpecialForm "ill-formed case expression: " form
Exercise 4
editLet's add string-length and string-ref:
primitives = [... ("string-length", stringLen), │ ("string-ref", stringRef), ...]
stringLen :: [LispVal] -> ThrowsError LispVal stringLen [(String s)] = Right $ Number $ fromIntegral $ length s stringLen [notString] = throwError $ TypeMismatch "string" notString stringLen badArgList = throwError $ NumArgs 1 badArgList
stringRef :: [LispVal] -> ThrowsError LispVal stringRef [(String s), (Number k)] | length s < k' + 1 = throwError $ Default "Out of bound error" | otherwise = Right $ String $ [s !! k'] where k' = fromIntegral k stringRef [(String s), notNum] = throwError $ TypeMismatch "number" notNum stringRef [notString, _] = throwError $ TypeMismatch "string" notString stringRef badArgList = throwError $ NumArgs 2 badArgList