Question

让我们考虑以下代码段：

blah :: a -> b -> a
blah x y = ble x where
    ble :: b -> b
    ble x = x

这在GHC下编译得很好，这实际上意味着第3行的b与第一行的b不同。

我的问题很简单：有没有办法以某种方式将ble的类型声明与外部上下文中使用的类型相关联，即blah的类型声明？

显然，这只是一个示例，而不是类型声明的真实用例。

Answer 1

这可以通过ScopedTypeVariables扩展程序实现。您需要使用显式的forall来将类型变量放入范围。

blah :: forall a b. a -> b -> a
blah x y = ble x where
    ble :: b -> b
    ble x = x

尝试在启用ScopedTypeVariables的情况下加载此定义会产生：

foo.hs:2:16:
    Couldn't match type `a' with `b'
      `a' is a rigid type variable bound by
          the type signature for blah :: a -> b -> a at foo.hs:2:1
      `b' is a rigid type variable bound by
          the type signature for blah :: a -> b -> a at foo.hs:2:1
    In the first argument of `ble', namely `x'
    In the expression: ble x
    In an equation for `blah':
        blah x y
          = ble x
          where
              ble :: b -> b
              ble x = x

您可以告诉GHC将两个b解释为相同的类型，因为错误表明a和b绑定在同一行。

Answer 2

如果您不想使用ScopedTypeVariables，则可以使用good ole fashion asTypeOf函数。

-- defined in Prelude
asTypeOf :: a -> a -> a
x `asTypeOf` y = x

blah :: a -> b -> a
blah x y = ble x where
    ble x = x `asTypeOf` y

当然，由于类型错误，这将无法编译。

更新

我想指出，有时你可能需要用asTypeOf来做你想做的事情。采用以下超级使用asTypeOf的示例，因为我不想考虑实际需要asTypeOf的案例。对于现实世界的案例，类似的解决方案也是如此。

foo :: Bounded a => Maybe a -> a
foo m = x
  where
    x = maxBound -- Q: how do I make (x :: a) when given (Maybe a)?
    _ = Just x `asTypeof` m -- A: witchcraft!

如何与外部语境中的类型相关联

2 个答案:

更新