Question

编辑：对不起，我的愚蠢。我正在使用7.6.3平台，但是阅读最新版本的在线文档，我就这样做了：

>:i Seq
newtype Seq a
  = Data.Sequence.Seq (Data.Sequence.FingerTree
                         (Data.Sequence.Elem a))
    -- Defined in `Data.Sequence'
instance Eq a => Eq (Seq a) -- Defined in `Data.Sequence'
instance Monad Seq -- Defined in `Data.Sequence'
instance Functor Seq -- Defined in `Data.Sequence'
instance Ord a => Ord (Seq a) -- Defined in `Data.Sequence'
instance Read a => Read (Seq a) -- Defined in `Data.Sequence'
instance Show a => Show (Seq a) -- Defined in `Data.Sequence'

并且看，没有应用实例。很抱歉打扰了，感谢bheklilr向我询问版本控制，我只是间隔开来。

以下让我感到困惑。肯定似乎有一个为Seq明确定义（＆lt; *＆gt;）的实例：

instance Applicative Seq where
pure = singleton
fs <*> xs = foldl' add empty fs
  where add ys f = ys >< fmap f xs

然而，这种情况发生了：

>:m + Data.Sequence  Control.Applicative 
>(*) <$> [1..4] <*> [3..4]
[3,4,6,8,9,12,12,16]
>(*) <$> fromList [1..4] <*> fromList [3..4]

<interactive>:58:25:
    No instance for (Applicative Seq) arising from a use of `<*>'
    Possible fix: add an instance declaration for (Applicative Seq)
    In the expression: (*) <$> fromList [1 .. 4] <*> fromList [3 .. 4]
    In an equation for `it':
        it = (*) <$> fromList [1 .. 4] <*> fromList [3 .. 4]
>:t (*) <$> fromList [1..4] 
(*) <$> fromList [1..4] :: (Enum a, Num a) => Seq (a -> a)

使用单例产生的输入略有不同（没有枚举限定），因此错误略有不同。

>:t (*) <$> singleton 3
(*) <$> singleton 3 :: Num a => Seq (a -> a)
>:t (*) <$> singleton 3 <*> fromList [3 .. 4]

<interactive>:1:21:
    Could not deduce (Applicative Seq) arising from a use of `<*>'
    from the context (Enum b, Num b)
      bound by the inferred type of it :: (Enum b, Num b) => Seq b
      at Top level
    Possible fix: add an instance declaration for (Applicative Seq)
    In the expression: (*) <$> singleton 3 <*> fromList [3 .. 4]

我认为Seq (a -> a)类型符合f (a -> b)所要求的(<*>) :: Applicative f => f (a -> b) -> f a -> f b。

我确信这很简单，我会因为询问而感到尴尬，但是事先谢谢。

Haskell Data.Sequence - 没有（Applicative Seq）的实例

0 个答案: