haskell - Haskell:有关部分申请的问题

我正在读《学习Haskell成就伟大》一书！由Miran Lipovaca撰写，并在第5章中了解高阶函数。

示例之一涉及以下功能：

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，我知道该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"

但是，我不理解第二个示例中串联运算符的工作方式。如何产生输出字符串“ HAHA HAHA HEY”？任何见解均表示赞赏。

最佳答案

对于第一个示例，我知道该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"

如果您从函数角度考虑，它会直接跳转到infix表达式（即在++之间）。

(++ " 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，则会得到

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 ("HAHA " ++) "HEY"执行相同的操作。

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

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