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同时匹配多个数据类型构造函数

haskell

Daniel · 技术社区 · 14 年前

假设我们有这个类型声明:

data D a = A a | B a | C a | D a | E a | F a

g x | x `is` [A,B,C] = 1
    | x `is` [D,E,F] = 2

而不是分别在每个构造函数上进行匹配。

6 回复 | 直到 14 年前

sth ACP 14 年前

如果您经常需要匹配同一组构造函数,那么helper函数可能是最简单的解决方案。例如:

getAbc :: D a -> Maybe a
getAbc (A v) = Just v
getAbc (B v) = Just v
getAbc (C v) = Just v
getAbc _     = Nothing

g 可以这样简化:

g x = g_ (getAbc x)
  where
    g_ (Just v) = 1
    g_ Nothing  = 2

或者,使用 maybe 功能:

g = maybe 2 (\v -> 1) . getAbc

kennytm 14 年前

编辑:如果所有构造函数都具有相同类型的字段,则可能会滥用函子:

{-# LANGUAGE DeriveFunctor #-}

data D a = A a | B a | C a | D a | E a | F a
    deriving (Eq, Functor)

isCons :: (Eq (f Int), Functor f) => f a -> (Int -> f Int) -> Bool
isCons k s = fmap (const 42) k == s 42

is :: (Eq (f Int), Functor f) => f a -> [Int -> f Int] -> Bool
is k l = any (isCons k) l

g :: D a -> Int
g x | x `is` [A,B,C] = 1
    | x `is` [D,E,F] = 2

{-# LANGUAGE DeriveDataTypeable #-}

import Data.Data

data D a = A a | B a | C a | D a | E a | F a
        deriving (Typeable, Data)

g :: Data a => D a -> Int
g x | y `elem` ["A","B","C"] = 1
    | y `elem` ["D","E","F"] = 2
    where y = showConstr (toConstr x)

ony 14 年前

我试着把“肯尼特”的答案概括为:

data D a = A a | B a | C a a | D
    deriving (Show, Eq, Functor)

class AutoBind a where
    bindSome :: forall b . (a -> b) -> b

instance AutoBind Bool where bindSome f = f False
instance Num a => AutoBind a where bindSome f = f 0

class AutoConst a b | a -> b where {- bind until target type -}
    bindAll :: a -> b

instance AutoBind a => AutoConst (a -> b) b where bindAll = bindSome
instance (AutoBind a, AutoConst b c) => AutoConst (a -> b) c where bindAll = bindAll . bindSome

isCons :: (Eq (f a), AutoBind a, AutoConst b (f a), Functor f) => f a -> b -> Bool
isCons x y = fmap (bindSome const) x == bindAll y

但由于某些原因它对构造函数不起作用 C

C. A. McCann Ravikant Cherukuri 14 年前

Data.Data 以及“占位符”类型?

{-# LANGUAGE DeriveDataTypeable #-}

import Data.Data 

data X = X deriving (Show, Data, Typeable)
data D a = A a | B a | C a | D a a | E a a | F a a
    deriving (Show, Data, Typeable)


matchCons :: (Data a) => D a -> [D X] -> Bool
matchCons x ys = any ((== xc) . toConstr) ys
    where xc = toConstr x

g :: (Data a) => D a -> Int
g x | matchCons x [A X, B X, C X] = 1
    | matchCons x [D X X, E X X, F X X] = 2

注意,这避免了类型签名/不同构造函数的问题。或许还有一种更清洁的方法来做类似的事情。

newacct 14 年前

我希望Haskell模式能够有一种方法来指定两个模式的“OR”,类似于 | 在OCaml中:

(* ocaml code *)
let g x = match x with
            A v | B v | C v -> 1
          | C v | D v | E v -> 2

gatoatigrado 13 年前

我也有同样的问题。我的解决方案是使用一个视图,尽管我个人希望使用更规范的语义等价的视图(在我正在编写的一些代码中,惰性保存非常关键,因此任何额外的不需要的模式匹配都可能使该技术不可用)。

{-# LANGUAGE ViewPatterns #-}

data D a = A a | B a | C a | D a | E a | F a
isABC (A v) = Just v
isABC (B v) = Just v
isABC (C v) = Just v
isABC _ = Nothing

f :: D Int -> Int
f (isABC -> Just v) = v
f (isABC -> Nothing) = 0