Haskell 高阶函数和结合性

Question

我正在学习 FP，在玩过 GHCi 之后有一些困惑。

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

分解求值 twice twice twice twice double 3(注意 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 的定义更改为 twice f a = f f a -- Equation 2，我应该将计算分解为左结合性，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)

按照您评论中的要求: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。