Haskell：关于部分应用的问题

我正在读一本书“为你学习哈斯克尔的伟大利益！”由Miran Lipovaca和第5章学习高阶函数

其中一个示例涉及以下功能：

applyTwice :: (a -> a) -> a -> a
applyTwice f x = f (f x)

以下是函数输出的示例：

ghci> applyTwice (++ " HAHA") "HEY"
"HEY HAHA HAHA"

ghci> applyTwice ("HAHA " ++) "HEY"
"HAHA HAHA HEY"

对于第一个示例，我了解字符串是通过以下方式使用串联运算符生成的：

"HEY" ++ " HAHA"
"HEY HAHA" ++ " HAHA"
"HEY HAHA HAHA"

"HEY" ++ " HAHA"
"HEY HAHA" ++ " HAHA"
"HEY HAHA HAHA"

但是，我不理解第二个示例中的串联运算符是如何工作的。输出字符串“哈哈嘿”是如何产生的？任何见解都将不胜感激

对于第一个示例，我了解字符串是通过以下方式使用串联运算符生成的：

"HEY" ++ " HAHA"
"HEY HAHA" ++ " HAHA"
"HEY HAHA HAHA"

"HEY" ++ " HAHA"
"HEY HAHA" ++ " HAHA"
"HEY HAHA HAHA"

与其直接跳转到中缀表达式（即，

++

介于两者之间），不如从函数的角度考虑

(++ " HAHA") :: [Char] -> [Char]   -- #1 this is a function (++ is partially applied)     
"HEY" :: [Char]

(++ " HAHA") "HEY"                 -- apply "HEY" as an argument to #1
-- same as "HEY" ++ " HAHA"

(+) :: (Num a) => a -> a -> a      -- #2 a binary function
(+) 1 2                            -- #3 apply 1 and 2 as arguments to #2
-- same as 1 + 2

-- technically, #3 is curried as
--    ((+) 1) 2                    -- i.e. (+) 1 is a partially applied function, which is then applied to 2     

如果将

（++“HAHA”）

替换为

applyTwice

的定义，您将得到

applyTwice f x = f (f x)
applyTwice (++ " HAHA") "HEY" = (++ " HAHA") ((++ " HAHA") "HEY")

                              = (++ " HAHA") ("HEY" ++ " HAHA")
                              = (++ " HAHA") ("HEY HAHA")
                              = "HEY HAHA" ++ " HAHA"
                              = "HEY HAHA HAHA"

现在对applyTwice做同样的操作（“哈哈”+）“嘿”

---

我是说。。。同样的方式？它变成了

“哈哈”+“嘿”

，然后变成

“哈哈”+“哈哈嘿”

，然后变成

“哈哈哈哈嘿”

。