一个Haskell函数，用于将单词列表转换为字符串

Question

例如：

wordsToString ["all","for","one","and","one","for","all"]
"all for one and one for all"

我的代码在没有类型声明的情况下工作：

wordsToString [] = ""
wordsToString [word] = word
wordsToString (word:words) = word ++ ' ':(wordsToString words)

但是当我进行类型检查时，它显示它是一个Chars列表，这对我来说似乎是错误的，因为我应该将输入声明为字符串列表并获取字符串作为输出：

*Main> :type wordsToString
wordsToString :: [[Char]] -> [Char]

我想将声明更改为wordsToString::[(String)]->[String]但不起作用

Answer 1

I want to change the declaration to wordsToString::[(String)]->[String] but it won't work

No, you want to change the declaration to wordsToString :: [String] -> String. You aren't getting a list of strings out, just a single one.

Answer 2

该函数名为concat：

concat :: Foldable t => t [a] -> [a]
concat xs = foldr (++) [] xs

在您的情况下，您希望在字符之间插入空格。此函数称为intercalate：

intercalate :: [a] -> [[a]] -> [a]

根据intersperse定义。