Question

假设：

import Data.Int (Int64)

data Unknown

class Meh a where
    meh :: a -> String
instance Meh Int64 where
    meh = show
instance Meh Unknown where
    meh _ = "Unknown"

很明显，meh只能使用Int64或Unknown。

但我们都知道

meh 1

会导致错误（文字“1”中没有（Num a0）的实例），因为1是Num a => a而编译器不知道Num的哪个实例应该是。

但是，（逻辑上）可以将1推断为Int64因为它传递给meh而meh只能传递Int64 1}}或Unknown。换句话说，应该可以计算Num和Meh的“交集”，并决定1是哪个实例。

所以我的问题是：你知道任何方式（编译器扩展，代码解决方法，任何东西）可以编写meh 1，其中1被正确推断为Int64没有添加类型签名或添加专门的功能？

Answer 1

在GHC 7.8.3中尝试您的示例时

>>> meh 1

我明白了：

<interactive>:2:1:
    No instance for (Meh a0) arising from a use of ‘meh’
    The type variable ‘a0’ is ambiguous
    Note: there are several potential instances:
      instance Meh Unknown -- Defined at Meh.hs:9:10
      instance Meh Int64 -- Defined at Meh.hs:7:10
    In the expression: meh 1
    In an equation for ‘it’: it = meh 1

<interactive>:2:5:
    No instance for (Num a0) arising from the literal ‘1’
    The type variable ‘a0’ is ambiguous
    Note: there are several potential instances:
    ...

这是因为编译器试图进行类型检查：

>>> meh (fromInteger (1 :: Integer))

你得到类似的错误：

<interactive>:4:1:
    No instance for (Meh a0) arising from a use of ‘meh’
    The type variable ‘a0’ is ambiguous
    Note: there are several potential instances:
      instance Meh Unknown -- Defined at Meh.hs:9:10
      instance Meh Int64 -- Defined at Meh.hs:7:10
    In the expression: meh (fromInteger (1 :: Integer))
    In an equation for ‘it’: it = meh (fromInteger (1 :: Integer))

<interactive>:4:6:
    No instance for (Num a0) arising from a use of ‘fromInteger’
    The type variable ‘a0’ is ambiguous
    Note: there are several potential instances:

类型类已打开。没有人限制别人定义：

newtype MyInt = MyInt Int deriving (Num)
instance Meh MyInt where
   meh = ...

然后即使在您的示例meh 1中也可以解析为(meh :: Int64 -> String) 1，通常它不能。{像“但基于范围中可见的Meh实例”这样的参数在Haskell中并不真正起作用，因为实例是全局的。

如果您定义data（已关闭），则可以使用{-# LANGUAGE OverloadedStrings #-}：

data Meh = MehInt Int64 | MehString String

instance IsString Meh where
   fromString = MehString

然后

"foo" :: Meh

将编译，因为GHC“将看到”

fromString "foo" :: Meh

同样，您可以定义不完整的Num实例，仅定义fromInteger。然而，我个人不喜欢不完整的实例，尽管在定义eDSL时这种情况很方便。

从上下文中猜测Num的正确实例

1 个答案: