为什么定义 Choice 的实例失败且值未知?

Why is defining an instance of Choice failing with unknown value?

更新:我在这里内联代码。

我正在尝试定义 Data.Profunctor.Choice 的实例,其中 right 是通过调用 left 定义的,但出于某种原因,编译器抱怨 left 未知.

newtype StateTrans s i o = ST (Tuple s i → Tuple s o)

instance stFunctor ∷ Functor (StateTrans s a) where
  map f (ST st) = ST $ second f <<< st

instance stProfunctor ∷ Profunctor (StateTrans s) where
  dimap f g (ST st) = ST $ second g <<< st <<< second f

instance stChoice ∷ Choice (StateTrans s) where
  left (ST f) = ST lf
    where
    lf (Tuple s (Left a)) = let (Tuple s' b) = f (Tuple s a)
                            in Tuple s' (Left b)
    lf (Tuple s (Right c)) = Tuple s (Right c)
  -- Fails to compile with: Unknown value left
  right f = arr mirror <<< left f <<< arr mirror
    where
    mirror Left x = Right x
    mirror Right x = Left x

可能是个愚蠢的错误,但我已经查看我的代码很长时间了,我无法弄清楚哪里出了问题。

(次要且不相关:在 leftRight 情况下,我必须解包并重新打包该值,以便它类型对齐。添加类型归属也无法编译.)

奇怪的是,我对 Strong 做同样的事情没有问题,请参阅:

instance stStrong ∷ Strong (StateTrans s) where
  first (ST f) = ST ff
    where
    ff (Tuple s (Tuple a c)) = let (Tuple s' b) = f $ Tuple s a
                               in Tuple s' (Tuple b c)
  second f = arr swap <<< first f <<< arr swap

首先,要使用 arr,您需要为 StateTrans s 声明 Category 实例:

instance stSemigroupoid :: Semigroupoid (StateTrans s) where
  compose (ST f1) (ST f2) = ST $ f1 <<< f2

instance stCategory :: Category (StateTrans s) where
  id = ST $ \x -> x

对于第二步,我需要添加更多类型注释(不确定为什么需要它们,但这样构建成功了):

choiceLeft :: forall input output a s. (StateTrans s) input output -> (StateTrans s) (Either input a) (Either output a)
choiceLeft (ST f) = ST lf
  where
    lf (Tuple s (Left a))  = let (Tuple s' b) = f (Tuple s a)
                             in Tuple s' (Left b)
    lf (Tuple s (Right c)) = Tuple s (Right c)

choiceRight :: forall input output t s. (StateTrans s) input output -> (StateTrans s) (Either t input) (Either t output)
choiceRight f = amirror <<< choiceLeft f <<< amirror
  where
    mirror :: forall a b. Either a b -> Either b a
    mirror (Left x)  = Right x
    mirror (Right x) = Left x
    amirror :: forall a b. StateTrans s (Either b a) (Either a b)
    amirror = arr mirror

instance stChoice ∷ Choice (StateTrans s) where
  left  = choiceLeft
  right = choiceRight

注意:使用 PureScript 版本 0.11.7purescript-profunctor 版本 3.2.0

我不是 100% 确定,因为粘贴的代码段不包含导入,但我怀疑您有 Data.Profunctor.Choice (class Choice) 而不是 Data.Profunctor.Choice (class Choice, left, right).

的导入

导入 class 不会隐式导入其成员,即使不这样做也可以在实例中定义它们。