Haskell 高阶函数和结合性

Question

我正在学习 FP，在玩过 GHCi 后有一些困惑。

假设我有 2 个简单的函数：

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

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

分解评价twice twice twice double 3（注意是3xtwice+1xdouble），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
-}

这是正确的吗？
根据this，如果我对twice的定义改为twice f a = f f a -- Equation 2，我应该分解求值，左结合性，如：

{-
   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
-}

对吗？

然而，最奇怪的是 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，函数应用是左关联的，所以：

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

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

根据您评论中的要求：请注意 twice returns 与其参数的类型相同。所以，twice twice 的类型就是 ((a -> a) -> (a -> a))

现在，让我们展开整个表达式：

(((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 double^16 3 == 2^16 * 3 如您所见。

Answer 2

假设n是一个自然数，g n定义如下，非正式地：

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

在你的情况下，twice f x = g 2 f x。

那么，可以证明

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

的确如此，

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”函数，然后重复 m 次以形成另一个函数，然后重复 m -再次，...每一步将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 双 16 倍，得到 3 * 2^16 = 196608.

Haskell 高阶函数和结合性

Haskell Higher Order Functions and Associativity

haskell

functional-programming

ghc

higher-order-functions