从上下文中猜测 Num 的正确实例

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只能取一个Int64或一个Unknown。

但我们都知道

meh 1

会导致错误（No instance for (Num a0) arising from the literal '1'）因为 1 是 Num a => a 并且编译器不知道 [=17= 的哪个实例] 应该是。

但是，（逻辑上）应该可以推断出 1 是一个 Int64，因为它被传递给 meh 而 meh 只能接受一个Int64 或 Unknown。换句话说，应该可以计算Num和Meh的"intersection"并决定1是哪个实例。

所以我的问题是：您是否知道有什么方法（编译器扩展、代码解决方法等）可以编写 meh 1，其中 1 被正确推断为 Int64 不添加类型签名或添加专用函数？

Answer 1

在 GHC 7.8.3 中尝试您的示例时

>>> meh 1

我得到：

<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:

Type-类已打开。没有人限制别人定义：

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

那么即使在你的例子中meh 1也可以解析为(meh :: Int64 -> String) 1，一般情况下是不行的。 "but based on the Meh instances visible in the scope" 之类的参数在 Haskell 中实际上不起作用，因为实例是全局的。

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

data Meh = MehInt Int64 | MehString String

instance IsString Meh where
   fromString = MehString

然后

"foo" :: Meh

将编译为 GHC "will see"

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，尽管这种情况在定义 eDSL 时很方便。

从上下文中猜测 Num 的正确实例

Guessing the correct instance of Num from the context

haskell

type-systems

typeclass