-- An Octet consists of eight bits. For our purposes, the most
-- significant bit will come "first." That is, b1 is in the 2^7
-- place while b8 is in the 2^0 place.
-data Octet = Octet { b1 :: Bit,
- b2 :: Bit,
- b3 :: Bit,
- b4 :: Bit,
- b5 :: Bit,
- b6 :: Bit,
- b7 :: Bit,
- b8 :: Bit }
- deriving (Eq, Show)
+data Octet = None | Octet { b1 :: Bit,
+ b2 :: Bit,
+ b3 :: Bit,
+ b4 :: Bit,
+ b5 :: Bit,
+ b6 :: Bit,
+ b7 :: Bit,
+ b8 :: Bit }
+ deriving (Eq)
+
+
+instance Show Octet where
+ show Octet.None = "None"
+ show oct = show (octet_to_int oct)
+
-- Convert each bit to its integer value, and multiply by the
-- appropriate power of two. Sum them up, and we should get an integer
8 * (bit_to_int (b5 x)) +
4 * (bit_to_int (b6 x)) +
2 * (bit_to_int (b7 x)) +
- 0 * (bit_to_int (b8 x))
+ 1 * (bit_to_int (b8 x))
+
--- Supply an integer greater than 255 at your own risk.
octet_from_int :: Int -> Octet
octet_from_int x
- | (x < 0) = min_octet
+ | (x < 0) || (x > 255) = Octet.None
| otherwise = (Octet a1 a2 a3 a4 a5 a6 a7 a8)
where
- a1 = if (x `mod` 128) > 0 then One else Zero
- a2 = if (x `mod` 64) > 0 then One else Zero
- a3 = if (x `mod` 32) > 0 then One else Zero
- a4 = if (x `mod` 16) > 0 then One else Zero
- a5 = if (x `mod` 8) > 0 then One else Zero
- a6 = if (x `mod` 4) > 0 then One else Zero
- a7 = if (x `mod` 2) > 0 then One else Zero
- a8 = if (x `mod` 1) > 0 then One else Zero
+ a1 = if (x > 128) then One else Zero
+ a2 = if ((x `mod` 128) >= 64) then One else Zero
+ a3 = if ((x `mod` 64) >= 32) then One else Zero
+ a4 = if ((x `mod` 32) >= 16) then One else Zero
+ a5 = if ((x `mod` 16) >= 8) then One else Zero
+ a6 = if ((x `mod` 8) >= 4) then One else Zero
+ a7 = if ((x `mod` 4) >= 2) then One else Zero
+ a8 = if ((x `mod` 2) == 1) then One else Zero
+
+
+octet_from_string :: String -> Octet
+octet_from_string s =
+ case (reads s :: [(Int, String)]) of
+ [] -> Octet.None
+ x:_ -> octet_from_int (fst x)
-- The octet with the least possible value.