[Haskell-cafe] Finally tagless - stuck with implementation of
"lam"
Jacques Carette
carette at mcmaster.ca
Mon Oct 5 12:06:46 EDT 2009
It's possible, but it's not nice. You need to be able to "get out of
the monad" to make the types match, i.e.
lam f = I (return $ \x -> let y = I (return x) in
unsafePerformIO $ unI (f y))
The use of IO 'forces' lam to transform its effectful input into an even
more effectful result. Actually, same goes for any monad used inside a
'repr'.
let_ and fix follow a similar pattern (although you can hide that
somewhat by re-using lam if you wish).
Jacques
Günther Schmidt wrote:
> Hi all,
>
> I'm playing around with finally tagless.
>
> Here is the class for my Syntax:
>
>
> class HOAS repr where
> lam :: (repr a -> repr b) -> repr (a -> b)
> app :: repr (a -> b) -> repr a -> repr b
> fix :: (repr a -> repr a) -> repr a
> let_ :: repr a -> (repr a -> repr b) -> repr b
>
> int :: Int -> repr Int
> add :: repr Int -> repr Int -> repr Int
> sub :: repr Int -> repr Int -> repr Int
> mul :: repr Int -> repr Int -> repr Int
>
>
> and here is one instance of that class for semantics.
>
>
>
> newtype I a = I { unI :: IO a }
>
> instance HOAS I where
> app e1 e2 = I (do
> e1' <- unI e1
> e2' <- unI e2
> return $ e1' e2')
> int i = I (putStrLn ("setting an integer: " ++ show i) >> return i)
> add e1 e2 = I (do
> e1' <- unI e1
> e2' <- unI e2
> putStrLn (printf "adding %d with %d" e1' e2')
> return $ e1' + e2')
> sub e1 e2 = I (do
> e1' <- unI e1
> e2' <- unI e2
> putStrLn (printf "subtracting %d from %d" e1' e2')
> return $ e1' - e2')
> mul e1 e2 = I (do
> e1' <- unI e1
> e2' <- unI e2
> putStrLn (printf "multiplying %d with %d" e1' e2')
> return $ e1' * e2')
>
>
> I'm stuck with the "lam" method, for 2 days now I've been trying to
> get it right, but failed.
>
> Is there a possibility that it isn't even possible?
>
> Günther
>
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