扣减类型在Haskell中失败。

Question

class (Eq a) => My a where
    f:: (a,a) -> (a,a)

instance My Int where
    f (a,b) = (a,b)

instance My Char where
    f (a,b) = (a,b)

和“这对情侣的专长”。它会产生编译错误，我不知道为什么。请帮助修复它，并解释为什么它是错误的。

instance (My a, My b) => My (a,b) where
    f ((a,b), (c,d)) = ( (f (a,b)), (f (c,d)) )

错误：

test.hs:11:31:
    Could not deduce (a ~ b)
    from the context (Eq (a, b), My a, My b)
      bound by the instance declaration at test.hs:10:10-33
      `a' is a rigid type variable bound by
          the instance declaration at test.hs:10:10
      `b' is a rigid type variable bound by
          the instance declaration at test.hs:10:10
    Expected type: (a, b)
      Actual type: (a, a)
    In the return type of a call of `f'
    In the expression: (f (a, b))
    In the expression: ((f (a, b)), (f (c, d)))
Failed, modules loaded: none.

Answer 1

错误信息是正确的。在您的元组实例中，

   f :: ((a,b), (a,b)) -> ((a,b), (a,b))

其中a和b在一般情况下是不同的类型。但是，对于任何实例，f只能对(a,a)元组进行操作，即您需要类型相等的a ~ b。因此

instance (My a) => My (a,a) where
  f (ab, cd) = (f ab, f cd)

会起作用..。或者，实际上，我记得这导致了一些问题(忘记了什么)，您最好将相等约束明确化。

instance (My a, a~b) => My (a,b) where

Answer 2

您可能需要以下实例：

instance (My a, My b) => My (a,b) where
    f ((a, b), (c, d)) = case (f (a, c), f (b, d)) of
      ((a1, a2), (b1, b2)) -> ((a1, b1), (a2, b2))

在(a, c) :: (a, a)和(b, d) :: (b, b)上面，所以我们将f应用于这些。我们得到一对类型的((a, a), (b, b))，我们重新排序，以获得((a, b), (a, b))。