Re: [Haskell-cafe] Thinking about an unlistN
On Sun, Aug 10, 2008 at 11:47 AM, Michael Feathers [EMAIL PROTECTED] wrote: unlist3 :: (a - a - a - b) - [a] - b unlist3 f (x:y:z:xs) = f x y z Oleg has written about this. Be careful, its easy to overdose on: Functions with the variable number of (variously typed) arguments http://okmij.org/ftp/Haskell/types.html#polyvar-fn I was also wondering whether the function can be generalized to N or whether this is just one of those edges in the type system that you can't abstract over. With template haskell I don't think that exists. Justin ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Thinking about an unlistN
I wrote this function the other day, and I was wondering if I'm missing something.. whether there is already a function or idiom around to do this. unlist3 :: (a - a - a - b) - [a] - b unlist3 f (x:y:z:xs) = f x y z I was also wondering whether the function can be generalized to N or whether this is just one of those edges in the type system that you can't abstract over. Thanks, Michael ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Thinking about an unlistN
I'm no expert, but it looks like the generalization of that would be some f that took a list: f :: [a] - b so what you'd have is a fold, right? foldr1 :: (a - a - a) - [a] - a Best, Philip Neustrom On Sun, Aug 10, 2008 at 11:47 AM, Michael Feathers [EMAIL PROTECTED] wrote: I wrote this function the other day, and I was wondering if I'm missing something.. whether there is already a function or idiom around to do this. unlist3 :: (a - a - a - b) - [a] - b unlist3 f (x:y:z:xs) = f x y z I was also wondering whether the function can be generalized to N or whether this is just one of those edges in the type system that you can't abstract over. Thanks, Michael ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Thinking about an unlistN
Philip, Thanks. It's not quite that, though. It's more like an adapter for a function with a specific arity. If I have, say, a function f :: a - a - a - a - b it would be nice to be able to just: unlistN 4 f [1..4] Michael (does look like there's no way to make that fly with the type system however) Philip Neustrom wrote: I'm no expert, but it looks like the generalization of that would be some f that took a list: f :: [a] - b so what you'd have is a fold, right? foldr1 :: (a - a - a) - [a] - a Best, Philip Neustrom On Sun, Aug 10, 2008 at 11:47 AM, Michael Feathers [EMAIL PROTECTED] wrote: I wrote this function the other day, and I was wondering if I'm missing something.. whether there is already a function or idiom around to do this. unlist3 :: (a - a - a - b) - [a] - b unlist3 f (x:y:z:xs) = f x y z I was also wondering whether the function can be generalized to N or whether this is just one of those edges in the type system that you can't abstract over. Thanks, Michael ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- Now Playing: Clammbon - 246 http://youtube.com/watch?v=PO77bN8W1mA ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Thinking about an unlistN
Michael Feathers wrote: I wrote this function the other day, and I was wondering if I'm missing something.. whether there is already a function or idiom around to do this. unlist3 :: (a - a - a - b) - [a] - b unlist3 f (x:y:z:xs) = f x y z I was also wondering whether the function can be generalized to N or whether this is just one of those edges in the type system that you can't abstract over. In Scheme and Lisp, the apply function does this, so the following Scheme definition would do the trick: (define unlist apply) (Although this doesn't ignore the xs as in the example; that would require some code to extract a sublist of the desired length.) In Haskell, bring on the type system shenanigans! ;) Anton ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Thinking about an unlistN
Hi,
On Sun, Aug 10, 2008 at 8:57 PM, Michael Feathers
[EMAIL PROTECTED] wrote:
If I have, say, a function f :: a - a - a - a - b it would be nice to be
able to just:
unlistN 4 f [1..4]
It indeed doesn't work like this; there's more than one way to do
something *like* this, if you really want to. The closest one is
probably to use type-level numbers:
{-# OPTIONS_GHC -fglasgow-exts #-}
data Zero
data Suc a
zero :: Zero; zero = undefined; suc :: a - Suc a; suc = undefined
one = suc zero; two = suc one; three = suc two; four = suc three
class Unlist n a b where
type UnlistFn n a b
unlist :: n - UnlistFn n a b - [a] - b
instance Unlist Zero a b where
type UnlistFn Zero a b = b
unlist _ r _ = r
instance Unlist n a b = Unlist (Suc n) a b where
type UnlistFn (Suc n) a b = a - UnlistFn n a b
unlist _ f (x:xs) = unlist (undefined :: n) (f x) xs
main = print (unlist four (,,,) abcd)
This prints ('a','b','c','d').
Hope this is fun[*],
- Benja
[*] I hesistate to say hope this helps in this case :-)
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