Question

例如，假设我们有以下数据结构：

data Foo = Bool Bool | Int Int | Double Double

现在，有更简单的方法吗？

foo :: Typeable a => a -> Foo
foo x = maybe (error "i dunno") id $
  liftM Bool   (cast x) `mplus`
  liftM Int    (cast x) `mplus`
  liftM Double (cast x)

有人想过在 Typeable 类型上为模式匹配制作语法吗？

Answer 1

使用typeOf和警卫：

foo x
    | tx == typeOf "str" = "string"
    | tx == typeOf True  = "bool"
    | otherwise          = "i dunno"
  where tx = typeOf x

Answer 2

你可以在这里使用视图模式，它们是非常方便的扩展。

{-# LANGUAGE ViewPatterns #-}

import Data.Data

data Foo = Bool Bool | Int Int | Double Double
         deriving (Show)

foo :: Typeable a => a -> Foo
foo (cast -> Just x) = Int x
foo (cast -> Just x) = Bool x
foo (cast -> Just x) = Double x
foo _ = error "i dunno"

结果：

*Main> :l foo_typeable.hs 
[1 of 1] Compiling Main             ( foo_typeable.hs, interpreted )
Ok, modules loaded: Main.
*Main> foo "123"
*** Exception: i dunno
*Main> foo 1
*** Exception: i dunno
*Main> foo (1 :: Int)
Int 1
*Main> foo (1 :: Integer)
*** Exception: i dunno
*Main> foo (1 :: Double)
Double 1.0

Answer 3

此版本不限于Bool，Int或Double，而String则为[Char]。

foo :: Typeable a => a -> String
foo = show . typeOf

可键入类型的“模式匹配”

3 个答案: