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