Write Yourself a Scheme in 48 Hours/Answers
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Chapter 1
[edit | edit source]Exercise 1
[edit | edit source]main :: IO ()
main = do args <- getArgs
putStrLn ("Hello, " ++ args!!0 ++ " " ++ args!!1)
Exercise 2
[edit | edit source]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
[edit | edit source]main :: IO ()
main = do putStrLn "What do they call thee at home?"
name <- getLine
putStrLn ("Ey up " ++ name)
Chapter 2
[edit | edit source]Section 3 - Parsing
[edit | edit source]Exercise 1
[edit | edit source]Part 1
[edit | edit source]parseNumber :: Parser LispVal
parseNumber = do x <- many1 digit
(return . Number . read) x
Part 2
[edit | edit source]In 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
[edit | edit source]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
[edit | edit source]escapedChars :: Parser Char
escapedChars = do char '\\'
x <- oneOf "\\\"nrt"
return $ case x of
'\\' -> x
'"' -> x
'n' -> '\n'
'r' -> '\r'
't' -> '\t'
Exercise 4
[edit | edit source]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 = 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
[edit | edit source]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
[edit | edit source]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
[edit | edit source]Ratio, 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
[edit | edit source]Exercise 1
[edit | edit source]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]
parseUnQuoteSplicing :: Parser LispVal
parseUnQuoteSplicing = do
char ','
char '@'
x <- parseExpr
return $ List [Atom "unquote-splicing", x]
Also add
<|> parseQuasiQuoted
<|> parseUnQuote
<|> parseUnQuoteSplicing
to parseExpr.
Exercise 2
[edit | edit source]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
[edit | edit source]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
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
[edit | edit source]Exercise 1
[edit | edit source]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 False listp _ = Bool False
Exercise 2
[edit | edit source]unpackNum :: LispVal -> Integer unpackNum (Number n) = n unpackNum _ = 0
Exercise 3
[edit | edit source]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
[edit | edit source]Exercise 1
[edit | edit source]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
[edit | edit source]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 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
[edit | edit source]cond
[edit | edit source]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
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
[edit | edit source]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
[edit | edit source]Let'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