在没有无关代码的情况下在Haskell中展开数据类型

Question

说我有

x = Just 2

是否有一种方法(最好是内置机制/函数)在单个语句中使用x,这样如果它是Just,则自动解包并使用2,如果它是Nothing,则引发异常？

那是,

(f x) + 2 == 4if x == Just 2,如果是,则引发异常x == Nothing.

Answer 1

Data.Maybe.fromJust 其他答案已经提到:

fromJust :: Maybe a -> a
fromJust Nothing  = error "Maybe.fromJust: Nothing"
fromJust (Just x) = x

还有maybe(在发现Prelude和Data.Maybe):

maybe :: b -> (a -> b) -> Maybe a -> b
maybe n _ Nothing  = n
maybe _ f (Just x) = f x

fromJust可以用以下方式编写maybe:

fromJust = maybe (error "Maybe.fromJust: Nothing") id

如您所见,maybe允许您灵活地处理两种情况而无需模式匹配:

\x -> maybe 0 (+ 2) x  -- Nothing -> 0, Just 2 -> 4

同样,Prelude并Data.Either有either :: (a -> c) -> (b -> c) -> Either a b -> c:

\x -> either (subtract 1) (* 2) x  -- Left 5 -> 4, Right 3 -> 6

如果定义数据类型

data MyDataType
  = TypeA { foo :: Int, bar :: String }
  | TypeB { foo :: Int,                baz :: () }
  | TypeC {             bar :: String, baz :: () }

像这样,你最终得到了访问器的部分功能.

foo :: MyDataType -> Int
bar :: MyDataType -> String
baz :: MyDataType -> ()

它们被称为部分函数,​​而不是总函数,因为它们只返回其输入子集的结果.

foo (TypeA { foo = 15, bar = "hello!" })  -- 15
bar (TypeB { foo = 12345679, baz = () })  -- error

Answer 2

对于这个特例,fromJust.一般来说

let Just k = x in f k + 2 == 4

此技巧适用于任何数据类型构造函数,并且非常常用(:)于非空列表.

Answer 3

只是应该做你想要的.