List 通过分发[]简化代码
我有以下功能List 通过分发[]简化代码,list,haskell,monads,List,Haskell,Monads,我有以下功能 type Assignment = [(Ref, Val)] predi2 :: [(Val,Val)] -> Ref -> Ref -> (Maybe Assignment -> [Maybe Assignment]) predi2 f k1 k2 = \ma -> case ma of Nothing -> [Nothing] Just a -> case lookup k1 a of Nothing
type Assignment = [(Ref, Val)]
predi2 :: [(Val,Val)] -> Ref -> Ref -> (Maybe Assignment -> [Maybe Assignment])
predi2 f k1 k2 = \ma -> case ma of
Nothing -> [Nothing]
Just a -> case lookup k1 a of
Nothing -> [Nothing]
Just e1 -> case lookup k2 a of
Nothing -> [Nothing]
Just e2 -> if elem (e1, e2) f
then [Just a]
else []
考虑到所有关于Maybe的例子,我想我可以在Maybe monad中把它简化为更简单的东西:
predi2 :: [(Val,Val)] -> Ref -> Ref -> (Maybe Assignment -> [Maybe Assignment])
predi2 f k1 k2 = [\ma -> do
a <- ma
e1 <- lookup k1 a
e2 <- lookup k2 a
if elem (e1, e2) f then (return a) else ???]
predi2::[(Val,Val)]->Ref->Ref->(可能分配->[可能分配])
预测2 f k1 k2=[\ma->do
不,不是
我将更像这样编写您的predi2
函数:
predi2 f k1 k2 ma = fromMaybe [Nothing] $ do
a <- ma
e1 <- lookup k1 a
e2 <- lookup k2 a
if elem (e1, e2) f then return [Just a] else return []
predi2f k1 k2 ma=fromMaybe[Nothing]$do
如果我正确理解这个问题,我认为这是可能的
让我印象深刻的是,您试图跟踪两种不同类型的失败,空列表表示最终的元素查找失败。可能
表示(1)原始可能分配
为无
或(2)lookup
失败。我们可以更明确地说明这一点。返回类型应该是
data FailureType = Existence | SeenInF
Either FailureType Assignment
然后,我们将使用模块将类型故障转换为或s
导入控制。错误
预测2::[(Val,Val)]->Ref->Ref->Maybe Assignment->任一故障类型赋值
predi2 f k1 k2 ma=执行a
import Control.Error
predi2 :: [(Val, Val)] -> Ref -> Ref -> Maybe Assignment -> Either FailureType Assignment
predi2 f k1 k2 ma = do a <- note Existence ma
e1 <- note Existence $ lookup k1 a
e2 <- note Existence $ lookup k2 a
note SeenInF $ guard $ (e1, e2) `elem` f
return a