haskell - 'where' 中的类型声明——这是怎么回事？

Question

在阅读 QuickCheck Manual 时，我遇到了以下示例:

prop_RevRev xs = reverse (reverse xs) == xs
  where types = xs::[Int]

手册接着说:

Properties must have monomorphic types. `Polymorphic' properties, such as the one above, must be restricted to a particular type to be used for testing. It is convenient to do so by stating the types of one or more arguments in a

where types = (x1 :: t1, x2 :: t2, ...)

clause. Note that types is not a keyword; this is just a local declaration which provides a convenient place to restrict the types of x1, x2 etc.

我以前从未在 Haskell 中看到过这样的技巧。这是我真正遇到的问题:

为什么这种类型声明的语法甚至存在？以下内容不能为我做什么？

prop_RevRev :: [Int] -> Bool
prop_RevRev xs = reverse (reverse xs) == xs

是否使用 where构成类型声明的“特殊”语法？或者它是一致的和合乎逻辑的(如果是这样，如何？)？

这是使用标准还是传统的 Haskell？

Answer 1

where不是类型声明的特殊语法。例如，这有效:

prop_RevRev :: [Int] -> Bool
prop_RevRev xs = ys == xs
  where ys = reverse (reverse xs)

这也是:

prop_RevRev xs = ys == xs
  where ys = reverse (reverse xs)
        ys :: [Int]

where type = (xs :: [Int])的优势|超过 prop_RevRev :: [Int] -> Bool是在后一种情况下您必须指定返回类型，而在前一种情况下编译器可以为您推断它。如果您有一个非 Boolean，这将很重要。属性，例如:

prop_positiveSum xs = 0 < length xs && all (0 <) xs ==> 0 < sum xs
  where types = (xs :: [Int])