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haskell - Haskell：关于部分应用的问题

问题描述

我正在阅读《Learn You a Haskell for Great Good!》这本书 由 Miran Lipovaca 撰写，并在第 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，我了解该字符串是通过以下方式使用连接运算符生成的：

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

但是，我不明白连接运算符在第二个示例中是如何工作的。输出字符串“HAHA HAHA HEY”是如何产生的？任何见解都值得赞赏。

标签： haskellfunctional-programminghigher-order-functionspartial-application

解决方案

对于第一个示例，我了解该字符串是通过以下方式使用连接运算符生成的：

"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），不如从函数的角度进行思考会有所帮助。

(++ " 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

如果你代(++ " HAHA")入 的定义applyTwice，你会得到

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".

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

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

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