Warning: file_get_contents(/data/phpspider/zhask/data//catemap/6/haskell/10.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Haskell 哈斯克尔可以';t从GADT构造函数中找出类型_Haskell_Gadt_Type Families - Fatal编程技术网

Haskell 哈斯克尔可以';t从GADT构造函数中找出类型

Haskell 哈斯克尔可以';t从GADT构造函数中找出类型,haskell,gadt,type-families,Haskell,Gadt,Type Families,对于这个冗长的例子,我事先表示歉意,我想不出一个较短的例子 让我们定义一个类型类框,它只包含另一个类型内容 {-# LANGUAGE GADTs #-} {-# LANGUAGE TypeFamilies #-} class Box t where type Content t data IntBox = IntBox data StringBox = StringBox 让我们写几个例子: instance Box

对于这个冗长的例子,我事先表示歉意,我想不出一个较短的例子

让我们定义一个类型类
,它只包含另一个类型
内容

{-# LANGUAGE GADTs                     #-}
{-# LANGUAGE TypeFamilies              #-}

class Box t where
    type Content t

data IntBox = IntBox
data StringBox = StringBox
让我们写几个例子:

instance Box IntBox where
    type Content IntBox = Int

instance Box StringBox where
    type Content StringBox = String

data JointBox a b = JointBox a b

instance (Box a, Box b) => Box (JointBox a b) where
    type Content (JointBox a b) = Either (Content a) (Content b)
到目前为止,所有这些都在编译和运行。进入GADTs。我想要一个代数数据类型,它由一个盒子和它的内容组成。构造函数完全决定了长方体的类型

data ABox t where
    AnIntBox :: IntBox -> ABox IntBox
    AStringBox :: StringBox -> ABox StringBox
    AnIntOrStringBox :: JointBox IntBox StringBox -> ABox (JointBox IntBox StringBox)
现在,在我看来,这意味着通过对
ABox
的构造函数进行模式匹配,应该确定框的类型及其内容。但事实似乎并非如此:

frobABox :: (Content t) ->  ABox t              -> IO ()
frobABox     int           (AnIntBox _)         =  print $ int + 3
frobABox     string        (AStringBox _)       =  putStrLn $ reverse string
frobABox    (Left int)     (AnIntOrStringBox _) =  print $ int + 6
frobABox    (Right string) (AnIntOrStringBox _) =  putStrLn $ reverse string
这带来了很多错误,其中这两个对我来说最重要:

GADTAndTypeClassBug.hs:29:14:
    Couldn't match expected type ‘Content t’
                with actual type ‘Either t0 t1’
    The type variables ‘t0’, ‘t1’ are ambiguous
    Relevant bindings include
      frobABox :: Content t -> ABox t -> IO ()
        (bound at GADTAndTypeClassBug.hs:27:1)
    In the pattern: Left int
    In an equation for ‘frobABox’:
        frobABox (Left int) (AnIntOrStringBox _) = print $ int + 6

GADTAndTypeClassBug.hs:30:14:
    Couldn't match expected type ‘Content t’
                with actual type ‘Either t2 t3’
    The type variables ‘t2’, ‘t3’ are ambiguous
    Relevant bindings include
      frobABox :: Content t -> ABox t -> IO ()
        (bound at GADTAndTypeClassBug.hs:27:1)
    In the pattern: Right string
    In an equation for ‘frobABox’:
        frobABox (Right string) (AnIntOrStringBox _)
          = putStrLn $ reverse string

GADTAndTypeClassBug.hs:30:71:
    Couldn't match expected type ‘[Char]’ with actual type ‘t3’
      ‘t3’ is untouchable
        inside the constraints (t ~ JointBox IntBox StringBox)
        bound by a pattern with constructor
                   AnIntOrStringBox :: JointBox IntBox StringBox
                                       -> ABox (JointBox IntBox StringBox),
                 in an equation for ‘frobABox’
        at GADTAndTypeClassBug.hs:30:29-46
    Relevant bindings include
      string :: t3 (bound at GADTAndTypeClassBug.hs:30:20)
    In the first argument of ‘reverse’, namely ‘string’
    In the second argument of ‘($)’, namely ‘reverse string’
没有类型族的更简单示例适用于:

data UnitOrEither t where
    AUnit :: () -> UnitOrEither ()
    AnEither :: Either String Int -> UnitOrEither (Either String Int)

frob :: UnitOrEither t -> IO ()
frob   (AUnit _)       =  print ()
frob   (AnEither _)    =  print "Either"

那么问题出在哪里呢?

对GADT模式匹配的细化从左到右进行。因此,通过对
frobABox
的第二个参数进行匹配而产生的类型细化将不适用于第一个参数

您可以通过翻转
frobABox
的参数来编译代码:

frobABox' :: ABox t              -> Content t -> IO ()
frobABox' (AnIntBox _)          int           =  print $ int + 3
frobABox' (AStringBox _)        string        =  putStrLn $ reverse string
frobABox' (AnIntOrStringBox _) (Left int)     =  print $ int + 6
frobABox' (AnIntOrStringBox _) (Right string) =  putStrLn $ reverse string

frobABox :: (Content t) ->  ABox t              -> IO ()
frobABox = flip frobABox'

你不需要一个类型类就可以得到一个类型族。您可以编写
类型族内容t;键入实例(JointBox a b)=(内容a)(内容b)
(对于
IntBox
/
StringBox
)中的任何一个。这是较少的代码,不会引入您不关心的类型级别名称。@DanielWagner,对,我只是简化了我正在使用的实际示例。