Haskell中的递归

Question

我还在学习Haskell,而且我正在做一些练习,但是我很伤心.所以我有一个名为"novel"的函数,它需要2个字符串和一个Int(novel :: (String, String, Int) -> String,它的参数).Novel的输入/输出必须如下所示:

> novel ("Rowling", "Harry Potter", 1998)
"Harry Potter (Rowling, 1998)"

这是我的新功能的代码,如上所述:

novel :: (String, String, Int) -> String
novel (author, book, year) = book ++ " (" ++ author ++ ", " ++ (show year) ++ ")" 

我正在尝试编写一个名为"cite"(cite :: [(String, String, Int)] -> String)的新函数.Cite的输入/输出应如下所示:

> cite [("author1", "book1", year1), ("author2", "book2", year2), ("author3", "book3", year3)]
"book1 (author1, year1)
book2 (author2, year2)
book3 (author3, year3)"

我试图递归地使用"小说",以获得所需的输出,但我不知道如何去做.

我尝试过的:

cite :: [(String, String, Int)] -> String                -- | Listed arguments
cite [] = ""                                             -- | Base Case
cite x:xs = [(novel (author, book, year)), (novel (author, book, year)), (novel (author, book, year))]

老实说,就我而言.显然,它不起作用,但我不知道该怎么做.

Answer 1

也许这会给你一个良好的开端:

cite :: [(String, String, Int)] -> String
cite [] = ""
cite (x:xs) = undefined -- put your code that recursively calls cite in here, hint: use ++ and "\n\"

模式匹配(x:xs)说明了这一点,给我列表中的第一项和列表x的尾部xs.这与写这个是一样的:

cite xs' = let x = head xs'
               xs = tail xs'
           in  undefined -- your code here

甚至

cite xs' = undefined -- your code here
    where
        x = head xs'
        xs = tail xs'

希望这有助于推动您朝着正确的方向前进.

编辑:OP询问如何递归执行此操作,下面是我原来的答案:

你可能应该重写你的基础案例cite [] = "".它并没有真正有所作为,但它有助于代码可读性.

让我们首先将":t map novel"放入ghci,看看你得到了什么:

> :t map novel
map novel :: [([Char], [Char], Int)] -> [[Char]]

我们可以改写为: map novel :: [(String, String, Int)] -> [String]

怎么样？因为map将一种类型转换a为另一种类型b并将其应用于列表中的每个项目.第一个参数map是任何带有一个参数的函数.究竟是什么呢novel.

但这并没有给我们你需要的东西,我们最终会得到一个字符串列表而不是字符串:

> cite [("author1", "book1", year1), ("author2", "book2", year2), ("author3", "book3", year3)]
["book1 (author1, year1)","book2 (author2, year2)","book3 (author3, year3)"]

并且您希望它是由换行符"\n"分隔的单个字符串.是否有一个函数可以获取字符串列表并将它们连接成一个字符串,但插入它们之间的分隔符？

首先让我们描述一下这样一个功能:String -> [String] -> String.接下来我们把它扔进Hoogle看看我们得到了什么:https://www.haskell.org/hoogle/ ?hoogle = String + - %3E+%5BString%5D+-%3E+String

啊,第二个功能intercalate听起来像我们需要的.它不仅适用于字符串,它适用于任何列表.它会如何工作？像这样的东西:

> import Data.List (intercalate)
> intercalate "\n" ["List","Of","Strings"]
"List\nOf\nStrings"

所以现在你可以结合插入和映射来获得你想要的东西.我将保留cite给你的定义.

编辑:完全忘了,实际上有一个专门的功能.如果您只是[String] -> String在Hoogle中搜索,您会发现unlines