Write Yourself a Scheme in 48 Hours/Answers


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List of Topics
 
List of Topics

 

Chapter 1

Exercise 1

main :: IO ()
main = do args <- getArgs
          putStrLn ("Hello, " ++ args!!0 ++ " " ++ args!!1)

Exercise 2

main :: 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

main :: IO ()
main = do putStrLn "What do they call thee at home?"
          name <- getLine
          putStrLn ("Ey up " ++ name)

Chapter 2

Section 3 - Parsing

Exercise 1

Part 1

parseNumber :: Parser LispVal
parseNumber = do x <- many1 digit
                (return . Number . read) x

Part 2

In order to anwer 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

We 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

escapedChars :: Parser Char
escapedChars = do char '\\' 
                  x <- oneOf "\\\"nrt" 
                  return $ case x of 
                    '\\' -> x
                    '"'  -> x
                    'n'  -> '\n'
                    'r'  -> '\r'
                    't'  -> '\t'

Exercise 4

First, 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 = parseDigital1 <|> parseDigital2 <|> parseHex <|> parseOct <|> parseBin

And the following new functions need to be added.


parseDigital1 :: Parser LispVal
parseDigital1 = many1 digit >>= (return . Number . read)
parseDigital2 :: Parser LispVal
parseDigital2 = 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

Exercise 5

data 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

A 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

Ratio, using Haskell's Ratio type:

 parseRatio :: Parser LispVal
 parseRatio = do x <- many1 digit
                 char '/'
                 y <- many1 digit
                 return $ Ratio ((read x) % (read y))

Additionally, import the 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) = 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 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

Exercise 1

These 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]

Also add

       <|> parseQuasiQuoted
       <|> parseUnQuote

to parseExpr.

Exercise 2

I 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

This 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


Alternative solution without a Nil constructor. 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

Exercise 1

Here 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 True
listp   _          = Bool False

Exercise 2

unpackNum :: LispVal -> Integer
unpackNum (Number n) = n
unpackNum _          = 0

Exercise 3

Add 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

Exercise 1

eval 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

Define 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 [l1@(List arg1), l2@(List arg2)] = eqvList eqv [l1, l2]

And add clauses for List and DottedList to the equal function:

 equal :: [LispVal] -> ThrowsError LispVal
 equal [l1@(List arg1), l2@(List arg2)] = eqvList equal [l1, l2]
 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

cond

Room 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


case

This 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

Last modified on 4 August 2013, at 05:05