在长长的例子中提前道歉,我无法想出更短的例子。
让我们定义一个类型类Box
,它只会包含另一种类型Content
。
{-# LANGUAGE GADTs #-}
{-# LANGUAGE TypeFamilies #-}
class Box t where
type Content t
data IntBox = IntBox
data StringBox = StringBox
让我们写几个实例:
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)
到目前为止所有编译和工作都是如此。输入GADT。我想要一个由一个盒子及其内容构成的代数数据类型。构造函数完全决定了框类型。
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ABox
的构造函数上的模式匹配,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
这引发了很多错误,其中这两个错误对我来说是最重要的:
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’
没有类型系列的更简单示例:
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"
那么问题是什么?
答案 0 :(得分:7)
GADT模式匹配的细化从左到右进行。因此,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
:
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'