x^y. Reply
Toralf Wittner [EMAIL PROTECTED] writes [..] data PowerNum = INT Integer | DBL Double deriving (Eq, Show, Read) [..] Now it basically works. However wouldn't it have been easier to write something like this: powPos :: Integer - Integer - Integer [..] powNeg :: Integer - Integer - Double [..] | y 0= 1 / fromInteger x * powNeg x (y + 1) Initially I wanted something as terse as the Python version, now I either have to write two functions or I need to define a type. Is there not an easier way to come as close as possible to the Python version? Thanks anyway - learnt a lot! For this particular task, the most natural solution is, probably, pow :: Fractional a = a - Integer - a pow a n = if n 0 then a^n else (1/a)^(- n) (^) is of standard, only pow adds the facility of negative n. Then, you need each time to convert the argument to appropriate type of Fractional: pow (fromInteger 2 :: Ratio.Rational) 2 -- 4 % 1 pow (fromInteger 2 :: Ratio.Rational) (-2) -- 1 % 4 pow (2 :: Double) (-2) -- 0.25 pow (2 :: Integer) (-2) -- ... No instance for (Fractional Integer) If you replace standard (and not lucky) Fractional with some class Foo (with multiplication mul and division div), make Integer an instance of Foo (where div may fail for some data), and program pow :: Foo a = a - Integer - a via mul, div, then it would work like this: pow (2 :: Integer) 2-- 4 pow (2 :: Integer) (-2) -- Error: cannot invert 2 :: Integer pow (2 :: Rational) (-2) -- 1%4 Another way is to try to straggle with overlapping instances by defining something like this: class Pow a b where pow :: a - Integer - b instance Num a = Pow a a where pow = (^) Fractional ? instance Num a = Pow Integer a where pow = ? If this succeeds, there will be also no need in new type constructors. - Serge Mechveliani [EMAIL PROTECTED] ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
x^y. Reply
Toralf Wittner [EMAIL PROTECTED] writes [..] power x y [..] | y 0 = x * power x (y-1) | y 0 = 1 / fromInteger x * power x (y+1) One recognizes that the function returns either an integer value if y 0 or a float value if y 0. Therefore I can't write a signature like pow :: Integer - Integer - Integer nor can I do pow :: Integer - Integer - Double. [..] How then would I write this function in Haskell (concerning types)? The type Rational fits the case n 0 too, and it includes Integer. But if you still need Integer | Double, you can, for example, introduce a new type of a disjoint union of the above two, and then, to compute like this: pow (Intg 2) 2 -- Intg 4 pow (Intg 2) (-2) -- D 0.25 pow (D 2.0) (-2) -- D 0.25 This is achieved by data PowerDom = Intg Integer | D Double deriving(Eq,Show) pow :: PowerDom - Integer - PowerDom pow x n = p x n where p (Intg m) n = if n 0 then Intg $ powerInteger m n else D $ powerDouble (fromInteger m :: Double) n p(D d) n = D $ powerDouble d n powerInteger m n = m^n :: Integer powerDouble :: Double - Integer - Double powerDoubled n = ... usual way for float - something like this. - Serge Mechveliani [EMAIL PROTECTED] ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: x^y. Reply
On Wednesday, 19. December 2001 09:12, S.D.Mechveliani wrote: [...] But if you still need Integer | Double, you can, for example, introduce a new type of a disjoint union of the above two, and then, to compute like this: pow (Intg 2) 2 -- Intg 4 pow (Intg 2) (-2) -- D 0.25 pow (D 2.0) (-2) -- D 0.25 This is achieved by data PowerDom = Intg Integer | D Double deriving(Eq,Show) pow :: PowerDom - Integer - PowerDom pow x n = p x n where p (Intg m) n = if n 0 then Intg $ powerInteger m n else D $ powerDouble (fromInteger m :: Double) n p(D d) n = D $ powerDouble d n powerInteger m n = m^n :: Integer powerDouble :: Double - Integer - Double powerDoubled n = ... usual way for float - something like this. This seems to be what I want. I tried it this way: module Main where import System main = do [a1, a2] - getArgs let x = read a1 let y = read a2 in putStrLn (show x ++ ^ ++ show y ++ = ++ show (pow x y)) data PowerNum = INT Integer | DBL Double deriving (Eq, Show, Read) pow :: PowerNum - Integer - PowerNum pow x y = z x y where z (INT x) y = if y 0 then INT $ powInteger x y else DBL $ powDouble (fromInteger x) y z (DBL x) y = DBL $ powDouble x y powInteger x y | x == 0= 0 | y == 0= 1 | y 0= x * powInteger x (y - 1) powDouble x y | x == 0= 0 | y == 0= 1 | y 0= 1 / x * powDouble x (y - 1) While GHC compiled this code I get a runtime error: Fail: Prelude.read: no parse and HUGS reports: ERROR: Illegal Haskell 98 class constraint in inferred type *** Expression : pow 1 2 *** Type : Num PowerNum = PowerNum Could you tell me what I did wrong? Thank you very much! Toralf - Serge Mechveliani [EMAIL PROTECTED] ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
