haskell - Rank 3 Types 函数示例

Question

我有这个 Rank 3 Type 函数定义：

f3 :: ((forall a. a -> a) -> Int) -> Bool -> Bool
f3 .... = ?

我正在努力为它写一个简单的例子。你能帮我吗？

Answer 1

f3 :: ((forall a. a -> a) -> Int) -> Bool -> Bool
f3 f b = (f id == 3) && b

f3_ex :: Bool
f3_ex = f3 f True where
   f :: (forall a. a -> a) -> Int
   f g = g 3

如评论中所述，将类型更改为更有趣：

f4 :: ((forall a. a -> a -> a) -> Int) -> Bool -> Bool
f4 f b = (f const == 3) && b

f4_ex :: Bool
f4_ex = f4 f True where
   f :: (forall a. a -> a -> a) -> Int
   f g = g 3 5


f5 :: ((forall a. a -> a -> a) -> Int) -> Bool -> Bool
f5 f b = f (if b then const else const id) == 42

f5_ex :: Bool
f5_ex = f5 f True where
   f :: (forall a. a -> a -> a) -> Int
   f g = g 3 5 + 39

这里f本质上是类型(Int, Int)，第一个组件由 给出，f const另一个由 给出f (const id)或等价f (flip const)。