6

通用编程时间！

如果我有一个功能：

f :: a1 -> a2 -> a3 -> ... -> an

和一个值

v :: aX   -- where 1 <= x < n

f在编译时不知道值的哪个参数v是正确的类型（如果有的话），我可以部分适用f于v吗？（使用 Typeable、Data、TH 或任何其他技巧）

稍微更扎实一点，我可以g在运行时构造函数（如下）吗？它实际上不必是多态的，我所有的类型都是单态的！

g :: (a1 -> a2 -> a3 -> a4 -> a5) -> a3 -> (a1 -> a2 -> a4 -> a5)
g f v = \x y z -> f x y v z

我知道，使用 Typeable （typeRepArgs特别是）v是 的第三个参数f，但这并不意味着我有办法部分应用f.

我的代码可能看起来像：

import Data.Typeable

data Box = forall a. Box (TyRep, a)

mkBox :: Typeable a => a -> Box
mkBox = (typeOf a, a)

g :: Box -> Box -> [Box]
g (Box (ft,f)) (Box (vt,v)) = 
    let argNums = [n | n <- [1..nrArgs], isNthArg n vt ft]
    in map (mkBox . magicApplyFunction f v) argNums

isNthArg :: Int -> TyRep -> TyRep -> Bool
isNthArg n arg func = Just arg == lookup n (zip [1..] (typeRepArgs func))

nrArgs :: TyRep -> Int
nrArgs = (\x -> x - 1) . length . typeRepArgs

有什么可以实现的magicApplyFunction吗？

编辑：我终于又开始玩这个了。魔术应用功能是：

buildFunc :: f -> x -> Int -> g
buildFunc f x 0 = unsafeCoerce f x
buildFunc f x i =
        let !res = \y -> (buildFunc (unsafeCoerce f y) x (i-1))
        in unsafeCoerce res
4

2 回答 2

2

我现在不打算在这里写整个解决方案，但我确信这可以完全用Data.Dynamicand来完成Typeable。来源dynApply并funResultTy应提供关键要素：

dynApply :: Dynamic -> Dynamic -> Maybe Dynamic
dynApply (Dynamic t1 f) (Dynamic t2 x) =
  case funResultTy t1 t2 of
    Just t3 -> Just (Dynamic t3 ((unsafeCoerce f) x))
    Nothing -> Nothing


funResultTy :: TypeRep -> TypeRep -> Maybe TypeRep
funResultTy trFun trArg
  = case splitTyConApp trFun of
      (tc, [t1,t2]) | tc == funTc && t1 == trArg -> Just t2
      _ -> Nothing

为了简单起见，我有type Box = (Dynamic, [Either TypeRep Dynamic]). 后者从参数的类型列表开始。magicApply将在框中查找第一个匹配的 TypeRep 并替换Dynamic该值。然后你可以有一个extract给定Box所有参数的 a ，实际执行dynApply调用以产生动态结果。

于 2011-04-21T15:14:15.793 回答
1

嗯..只能打字？好的旧的 OverlappingInstances 怎么样？

{-# LANGUAGE MultiParamTypeClasses, FlexibleInstances, TypeFamilies,
UndecidableInstances, IncoherentInstances, ScopedTypeVariables #-}

class Magical a b c where
    apply :: a -> b -> c

instance (AreEqual a c e, Magical' e (a -> b) c r) => Magical (a -> b) c r where
    apply f a = apply' (undefined :: e) f a


class Magical' e a b c where
    apply' :: e -> a -> b -> c

instance (r ~ b) => Magical' True (a -> b) a r where
    apply' _ f a = f a

instance (Magical b c d, r ~ (a -> d)) => Magical' False (a -> b) c r where
    apply' _ f c = \a -> apply (f a) c


data True
data False

class AreEqual a b r
instance (r ~ True) => AreEqual a a r
instance (r ~ False) => AreEqual a b r


test :: Int -> Char -> Bool
test i c = True

t1 = apply test (5::Int)
t2 = apply test 'c'
于 2011-04-21T15:45:10.713 回答