是否可以在quickcheck中生成任意函数

我试着写一个快速的身份检查测试

f $ y = f y

我最初的计划是编写一个任意的生成器,它返回具有签名的函数&integer Gen (Int -> Int, Int)

在 prop_DollerDoesNothing 使用/不使用 $ 得到同样的结果。

这是我的密码:

  prop_DollarDoesNothing :: Property
  prop_DollarDoesNothing =
    forAll arbitraryFuncInt (\(f, y) -> (f $ y) == (f y))

  arbitraryFuncInt :: Gen (Int -> Int, Int)
  arbitraryFuncInt = do
    f <- elements [(\x -> x*2), (\x -> x+3), (\x -> x-2)]
    y <- arbitrary :: Gen Int
    return (f, y)

它生成了以下有用的错误消息:

    * No instance for (Show (Int -> Int))
        arising from a use of `forAll'
        (maybe you haven't applied a function to enough arguments?)
    * In the expression:
        forAll arbitraryFuncInt (\ (f, y) -> (f $ y) == (f y))
      In an equation for `prop_DollarDoesNothing':
          prop_DollarDoesNothing
            = forAll arbitraryFuncInt (\ (f, y) -> (f $ y) == (f y))

所以,我修正了错误,通过应用任意函数并返回 arbitraryFuncInt

  prop_DollarDoesNothing :: Property
  prop_DollarDoesNothing =
    forAll arbitraryFuncInt (\(x, y) -> x == y)

  arbitraryFuncInt :: Gen (Int, Int)
  arbitraryFuncInt = do
    f <- elements [(\x -> x*2), (\x -> x+3), (\x -> x-2)]
    y <- arbitrary :: Gen Int
    return (f $ y, f y)

我的问题是:

1. 是否由于没有 Show ?
2. 我能为 Show (Int -> Int) 使 # 1 可能吗?
3. 可以在给定类型签名的情况下快速检查生成任意函数,在这种情况下,我将测试所有函数(给定类型)的真实身份。上面,我手动指定了3个测试函数,我希望以某种方式实现自动化,理想情况下是这样 f <- arbitrary :: Gen (Int -> Int)