Haskell Quiz/Bytecode Compiler/Solution Justin Bailey

This solution should work correctly. All test strings from the quiz evaluate to the correct values. To see it for yourself, execute the interpret_tests function.

To see the (symbolic) byte codes generated, run generate_tests. To see the actual byte codes, run compile_tests. To see that the values produced by each expression match those expected, run eval_tests. The last actually evaluates the AST, without generating any bytescodes. The tests are contained in the variables test1,test2, ..., test6, which correspond to the six "test_n" methods found in the quiz's test program.

The byte codes aren't optimized. For example, SWAP is never used. However, they should produce correct results (even for negative and LCONST/CONST values).

The code below is literate Haskell.

\begin{code}
import Text.ParserCombinators.Parsec hiding (parse)
import qualified Text.ParserCombinators.Parsec as P (parse)
import Text.ParserCombinators.Parsec.Expr
import Data.Bits
import Data.Int

-- Represents various operations that can be applied
-- to expressions.
data Op = Plus | Minus | Mult | Div | Pow | Mod | Neg
  deriving (Show, Eq)

-- Represents expression we can build - either numbers or expressions
-- connected by operators. This structure is the basis of the AST built
-- when parsing
data Expression = Statement Op Expression Expression
           | Val Integer
           | Empty
  deriving (Show)

-- Define the byte codes that can be generated. 
data Bytecode = NOOP | CONST Integer | LCONST Integer
            | ADD
            | SUB
            | MUL
            | POW
            | DIV
            | MOD
            | SWAP
  deriving (Show)


-- Using imported Parsec.Expr library, build a parser for expressions.
expr :: Parser Expression
expr =
  buildExpressionParser table factor
  <?> "expression"
  where
  -- Recognizes a factor in an expression
  factor  =
    do{ char '('
          ; x <- expr
          ; char ')'
          ; return x 
          }
      <|> number
      <?> "simple expression"
  -- Recognizes a number
  number  :: Parser Expression 
  number  = do{ ds <- many1 digit
              ; return (Val (read ds))
              }
          <?> "number"
  -- Specifies operator, associativity, precendence, and constructor to execute
  -- and built AST with.
  table =
    [[prefix "-" (Statement Mult (Val (-1)))],
      [binary "^" (Statement Pow) AssocRight],
      [binary "*" (Statement Mult) AssocLeft, binary "/" (Statement Div) AssocLeft, binary "%" (Statement Mod) AssocLeft],
      [binary "+" (Statement Plus) AssocLeft, binary "-" (Statement Minus) AssocLeft]
       ]          
    where
      binary s f assoc
         = Infix (do{ string s; return f}) assoc
      prefix s f 
         = Prefix (do{ string s; return f})

-- Parses a string into an AST, using the parser defined above
parse s = case P.parse expr "" s of
  Right ast -> ast
  Left e -> error $ show e
  
-- Take AST and evaluate (mostly for testing)
eval (Val n) = n
eval (Statement op left right)
        | op == Mult = eval left * eval right
        | op == Minus = eval left - eval right
        | op == Plus = eval left + eval right
        | op == Div = eval left `div` eval right
        | op == Pow = eval left ^ eval right
        | op == Mod = eval left `mod` eval right

-- Takes an AST and turns it into a byte code list
generate stmt = generate' stmt []
       where
               generate' (Statement op left right) instr =
                       let
                               li = generate' left instr
                               ri = generate' right instr
                               lri = li ++ ri
                       in case op of
                               Plus -> lri ++ [ADD]
                               Minus -> lri ++ [SUB]
                               Mult -> lri ++ [MUL]
                               Div -> lri ++ [DIV]
                               Mod -> lri ++ [MOD]
                               Pow -> lri ++ [POW]
               generate' (Val n) instr =
                if abs(n) > 32768
                then LCONST n : instr  
                else CONST n : instr

-- Takes a statement and converts it into a list of actual bytes to
-- be interpreted
compile s = toBytes (generate $ parse s)

-- Convert a list of byte codes to a list of integer codes. If LCONST or CONST
-- instruction are seen, correct byte representantion is produced
toBytes ((NOOP):xs) = 0 : toBytes xs 
toBytes ((CONST n):xs) = 1 : (toConstBytes (fromInteger n)) ++ toBytes xs
toBytes ((LCONST n):xs) = 2 : (toLConstBytes (fromInteger n)) ++ toBytes xs
toBytes ((ADD):xs) = 0x0a : toBytes xs
toBytes ((SUB):xs) = 0x0b : toBytes xs
toBytes ((MUL):xs) = 0x0c : toBytes xs
toBytes ((POW):xs) = 0x0d : toBytes xs
toBytes ((DIV):xs) = 0x0e : toBytes xs
toBytes ((MOD):xs) = 0x0f : toBytes xs
toBytes ((SWAP):xs) = 0x0a : toBytes xs
toBytes [] = []

-- Convert number to CONST representation (2 element list)
toConstBytes n = toByteList 2 n 
toLConstBytes n = toByteList 4 n 

-- Convert a number into a list of 8-bit bytes (big-endian/network byte order).
-- Make sure final list is size elements long
toByteList ::  Bits Int => Int -> Int -> [Int]
toByteList size n = reverse $ take size (toByteList' n)
    where
      toByteList' a = (a .&. 255) : toByteList' (a `shiftR` 8)

-- All tests defined by the quiz, with the associated values they should evaluate to.
test1 = [(2+2, "2+2"), (2-2, "2-2"), (2*2, "2*2"), (2^2, "2^2"), (2 `div` 2, "2/2"),
  (2 `mod` 2, "2%2"), (3 `mod` 2, "3%2")]

test2 = [(2+2+2, "2+2+2"), (2-2-2, "2-2-2"), (2*2*2, "2*2*2"), (2^2^2, "2^2^2"), (4 `div` 2 `div` 2, "4/2/2"),
  (7`mod`2`mod`1, "7%2%1")]
  
test3 = [(2+2-2, "2+2-2"), (2-2+2, "2-2+2"), (2*2+2, "2*2+2"), (2^2+2, "2^2+2"),
  (4 `div` 2+2, "4/2+2"), (7`mod`2+1, "7%2+1")]

test4 = [(2+(2-2), "2+(2-2)"), (2-(2+2), "2-(2+2)"), (2+(2*2), "2+(2*2)"), (2*(2+2), "2*(2+2)"),
  (2^(2+2), "2^(2+2)"), (4 `div` (2+2), "4/(2+2)"), (7`mod`(2+1), "7%(2+1)")]

test5 = [(-2+(2-2), "-2+(2-2)"), (2-(-2+2), "2-(-2+2)"), (2+(2 * -2), "2+(2*-2)")]

test6 = [((3 `div` 3)+(8-2), "(3/3)+(8-2)"), ((1+3) `div` (2 `div` 2)*(10-8), "(1+3)/(2/2)*(10-8)"), 
    ((1*3)*4*(5*6), "(1*3)*4*(5*6)"), ((10`mod`3)*(2+2), "(10%3)*(2+2)"), (2^(2+(3 `div` 2)^2), "2^(2+(3/2)^2)"),
    ((10 `div` (2+3)*4), "(10/(2+3)*4)"), (5+((5*4)`mod`(2+1)), "5+((5*4)%(2+1))")]

-- Evaluates the tests and makes sure the expressions match the expected values
eval_tests = concat $ map eval_tests [test1, test2, test3, test4, test5, test6]
  where
    eval_tests ((val, stmt):ts) =
      let eval_val = eval $ parse stmt 
      in
        if val == eval_val 
        then ("Passed: " ++ stmt) : eval_tests ts
        else ("Failed: " ++ stmt ++ "(" ++ show eval_val ++ ")") : eval_tests ts
    eval_tests [] = []

-- Takes all the tests and displays symbolic bytes codes for each
generate_tests = concat $ map generate_all [test1,test2,test3,test4,test5,test6]
  where generate_all ((val, stmt):ts) = (stmt, generate (parse stmt)) : generate_all ts
        generate_all [] = []
        
-- Takes all tests and generates a list of bytes representing them
compile_tests = concat $ map compile_all [test1,test2,test3,test4,test5,test6]
  where compile_all ((val, stmt):ts) = (stmt, compile stmt) : compile_all ts
        compile_all [] = []

interpret_tests = concat $ map f' [test1, test2, test3, test4, test5, test6]
  where
    f' tests = map f'' tests
    f'' (expected, stmt) =
      let value = fromIntegral $ interpret [] $ compile stmt
      in
        if value == expected
        then "Passed: " ++ stmt
        else "Failed: " ++ stmt ++ "(" ++ (show value) ++ ")"

fromBytes n xs =
  let int16 = (fromIntegral ((fromIntegral int32) :: Int16)) :: Int
      int32 = byte xs
      byte xs = foldl (\accum byte -> (accum `shiftL` 8) .|. (byte)) (head xs) (take (n - 1) (tail xs))
  in
    if n == 2
    then int16 
    else int32 
   
interpret [] [] = error "no result produced"
interpret (s1:s) [] = s1
interpret s (o:xs) | o < 10 = interpret ((fromBytes (o*2) xs):s) (drop (o*2) xs)
interpret (s1:s2:s) (o:xs)
  | o == 16 = interpret (s2:s1:s) xs
  | otherwise = interpret (((case o of 10 -> (+); 11 -> (-); 12 -> (*); 13 -> (^); 14 -> div; 15 -> mod) s2 s1):s) xs
 
\end{code}