Haskell 高阶函数和关联性

Question

我正在学习 FP，在玩过 GHCi 之后有些困惑。

假设我有 2 个简单的功能：

twice :: (a -> a) -> (a -> a)
twice f a = f (f a) -- Equation 1

double :: Int -> Int
double = \x -> x * 2

将评估分解twice twice twice double 3 （注意 3x twice +1x double ），我会：

{-
   twice twice twice double 3
== (twice twice twice double) 3
== (twice twice (twice double)) 3
== (twice (twice (double double 3))) 
== (twice ((double double) (double double 3))) 
== (((double double) (double double)) ((double double) (double double 3))) 
== 768
-}

• 这个对吗？
• 据此，如果我对twice的定义更改为twice fa = ffa -- Equation 2 ，我应该将评估分解为左结合性，如下：

{-
   twice (twice twice double) 3
== (twice twice double) (twice twice double) 3
== ((twice double)(twice double)) ((twice double)(twice double)) 3
== ((double double)(double double)) ((double double)(double double)) 3
== (double (double (double (double (double (double (double (double 3 ) ) ) ) ) ) )
== 768
-}

正确的？

• 然而，最奇怪的部分是 GHC REPL 给了我196608 (2^16*3) 的答案：

> twice twice twice double 3
196608

这让我很困惑。 我会在哪里犯错？ 谢谢。

Answer 1

正如评论所说，function 应用程序是左关联的，所以：

twice twice twice double 3 == (((twice twice) twice) double) 3

which is not the same as:     twice (twice twice double 3)

根据您的评论要求：请注意， twice返回相同类型的参数。 所以， twice twice的类型就是((a -> a) -> (a -> a))

现在，让我们扩展整个表达式：

(((twice twice) twice) double) 3 ==> ((twice (twice twice)) double) 3
                                 ==> (((twice twice) ((twice twice) double)) 3
                                 ==> (twice (twice ((twice twice) double))) 3
                                 ==> (twice (twice (twice (twice double)))) 3

twice double ==> double^2
twice (twice double) ==> double^4
twice (twice (twice double)) ==> double^8
twice (twice (twice (twice double))) == double^16

和你发现的double^16 3 == 2^16 * 3 。

Answer 2

假设n是一个自然数，并且gn定义如下，非正式地：

g n = \f x -> f (f (f (.... (f x))))  -- n times f on x

在您的情况下， twice fx = g 2 fx 。

那么，可以证明

g n (g m) f x = g (m^n) f x

的确，

g n (g m) f x =
(g m (g m (g m (g m .... (g m f))))) x =  -- n times (g m) on f

所以它采用“ m次重复f ” function，然后重复该m次以形成另一个 function，然后再重复该m次，...每一步都将f的应用次数乘以m ，所以我们得到m^n 。

回到你的案例

twice twice twice double 3 =
g 2 (g 2) (g 2) double 3 =
g (2^2) (g 2) double 3 =
g (2^(2^2)) double 3 =
double (double (double .... 3)) -- 2^(2^2) = 2^4 = 16 times

所以，我们得到3倍16倍，得到3 * 2^16 = 196608 。