Hi Patrick, thanks for taking your time! On Fri, May 6, 2011 at 18:45, Patrick LeBoutillier wrote:

I don't see how you can make the StackBool something else than Bool. The CMP and BRANCH operators seem to me very Boolean by definition. Can you provide an example of how it would work?

Patrick

I actually can :) I've been hacking today, and the result is: {-# LANGUAGE MultiParamTypeClasses #-} {-# LANGUAGE FunctionalDependencies #-} class (Show a) => AbsInteger a b | a -> b where (+) :: a -> a -> a (==) :: a -> a -> b absInteger :: Integer -> a class (Show a) => AbsBool a where cond :: a -> [Code] -> [Code] -> [Code] absBool :: Bool -> a instance AbsBool Bool where cond b c1 c2 = if b then c1 else c2 absBool b = b instance AbsInteger Integer Bool where a + b = (Prelude.+) a b a == b = (Prelude.==) a b absInteger a = a Now, if I would like to use my own integer type or my own bool type, I can do: data MyInteger = MyInteger Integer deriving (Show) data MyBool = MyBool Bool deriving (Show) instance AbsBool MyBool where cond (MyBool b) c1 c2 = if b then c1 else c2 absBool b = MyBool b instance AbsInteger MyInteger MyBool where (MyInteger a) + (MyInteger b) = MyInteger $ (Prelude.+) a b (MyInteger a) + (MyInteger b) = MyBool $ (Prelude.==) a b absInteger a = MyInteger a The eval function has to be tweaked a little bit, but it is almost plug'n'play :) However, due to the functional dependency of AbsInteger, this does not allow me to use MyInteger and Bool, since there can only be one pair of an integer type and a bool type, and Integer and Bool are already instantiating AbsInteger . Do you know some of way of making it possible to have several integer type and boolean type pairs instantiating the class AbsInteger?

Erik, On Fri, May 6, 2011 at 1:39 PM, Erik Helin wrote:

Hi Patrick, thanks for taking your time!

On Fri, May 6, 2011 at 18:45, Patrick LeBoutillier wrote:

...

I don't see how you can make the StackBool something else than Bool. The CMP and BRANCH operators seem to me very Boolean by definition. Can you provide an example of how it would work?

Patrick

I actually can :) I've been hacking today, and the result is:

Ok, now I get it! Did you try Brent's suggestion? For me it worked great and also allows you to drop the language extentions. Patrick -- ===================== Patrick LeBoutillier Rosemère, Québec, Canada

Erik, Here's the code I came up with. One thing I changed was using the names "add" and "eq" for the AbsInteger type class (I got confused by all the =s and +s). Coming from very imperative (and mostly dynamically typed) background I must admit that this kind of stuff blows my mind a bit... Patrick ========================== data Code a = PUSH a | ADD | CMP | BRANCH [Code a] [Code a] deriving (Show) data StackElement a b = StackValue a | StackBool b deriving (Show) type Stack a b = [StackElement a b] class AbsInteger i where add :: i -> i -> i eq :: AbsBool b => i -> i -> b absInteger :: Integer -> i class AbsBool b where cond :: (AbsInteger i) => b -> [Code i] -> [Code i] -> [Code i] absBool :: Bool -> b eval :: (AbsInteger a, AbsBool b) => [Code a] -> Stack a b -> Stack a b eval [] s = s eval (PUSH n:t) s = eval t (StackValue n:s) eval (ADD:t) (StackValue x:StackValue y:s) = eval t (StackValue (x `add` y):s) eval (CMP:t) (StackValue x:StackValue y:s) = eval t (StackBool (x `eq` y):s) eval (BRANCH c1 c2:t) (StackBool b:s) = eval ((cond b c1 c2) ++ t) s instance AbsBool Bool where cond b c1 c2 = if b then c1 else c2 absBool b = b instance AbsInteger Integer where a `add` b = (+) a b a `eq` b = absBool $ a == b absInteger a = a data MyInteger = MyInteger Integer deriving (Show) data MyBool = MyBool Bool deriving (Show) instance AbsBool MyBool where cond (MyBool b) c1 c2 = if b then c1 else c2 absBool b = MyBool b instance AbsInteger MyInteger where (MyInteger a) `add` (MyInteger b) = MyInteger $ (+) a b (MyInteger a) `eq` (MyInteger b) = absBool $ (==) a b absInteger a = MyInteger a code :: AbsInteger a => [Code a] code = [PUSH . absInteger $ 1, PUSH . absInteger $ 2, ADD, PUSH . absInteger $ 3, CMP, BRANCH [PUSH . absInteger $ 1] [PUSH . absInteger $ 0]] test1 = eval code [] :: Stack Integer Bool test2 = eval code [] :: Stack Integer MyBool test3 = eval code [] :: Stack MyInteger Bool test4 = eval code [] :: Stack MyInteger MyBool On Fri, May 6, 2011 at 3:32 PM, Erik Helin wrote:

On Fri, May 6, 2011 at 20:01, Brent Yorgey wrote:

...

Why not use

...

class (Show a) => AbsInteger a where     (+)        :: a -> a -> a     (==)       :: AbsBool b => a -> a -> b     absInteger :: Integer -> a

?

Because I believe (I might be very wrong now, I am very new to Haskell) that I would run into problem with the b in "AbsBool b => a -> a -> b" being unbounded when it would be used in:

data StackElement a b = StackInteger a                                   |  StackBool b                                   deriving (Show)

eval :: (AbsInteger a, AbsNum b) => [Code] -> [StackElement a b] -> [StackElement a b] {- some cases omitted -} eval (CMP:c) (StackInteger x:StackInteger y:t) = eval c (StackBool (x == y):t)

Now, GHC tells me that the type b returned by x == y does not need to equal the b specified in the type definition of eval.

Did I do something wrong in my definiton of eval?

On Fri, May 6, 2011 at 21:12, Patrick LeBoutillier wrote:

...

Did you try Brent's suggestion? For me it worked great and also allows you to drop the language extentions.

I tried, but I didn't manage to get it working, due to the what I describe above. Could you post your code using Brent's suggestion?

Clearly I am doing something wrong here, I just don't know what it is...

-- ===================== Patrick LeBoutillier Rosemère, Québec, Canada

On Saturday 07 May 2011 11:39:44, Erik Helin wrote:

Thank you Patrick for posting your code, it helps a lot too see a complete example!

On Sat, May 7, 2011 at 03:11, Patrick LeBoutillier

wrote:

...

instance AbsInteger Integer where a `add` b = (+) a b a `eq` b = absBool $ a == b absInteger a = a

In my code, I did not use "a `eq` b = absBool $ a == b". Instead, I used "a `eq` b = a == b". I thought, since "a == b" is of type Bool, and Bool is an instance of AbsBool, I can just return type Bool.

Why isn't this the case? What is it about typeclasses that I'm not understanding?

A type signature foo :: (Bar b) => SomeType -> b says "given a value of SomeType, I can produce return values of *any* type, as long as it's an instance of Bar". The caller decides which type it wants and the callee must be able to produce it. In OO, on the other hand, the callee decides which type it returns, a signature Bar foo(SomeType s); (where Bar is an interface) says foo will return values of a type, all you know about which is that it implements Bar. Using explicit quantification, the Haskell type is foo :: forall b. (Bar b) => SomeType -> b while the OO-type would correspond to foo :: exists b. (Bar b) => SomeType -> b (the latter type doesn't exist in Haskell).