我需要创建自己的单词函数。它需要字符串并放入有空间的列表中。例如,字符串“我需要帮助”将导致[“i”,“need”,“help”]。 定义必须完全
anything :: String -> [String]
我目前想出的愚蠢的解决方案看起来像这样(也不起作用)
test :: String -> [String]
test d = beforep d : (test (afterp d)) : []
beforep :: String -> String
beforep d = takeWhile (/=' ') d
afterp :: String -> String
afterp d = if (dropWhile (/=' ') d)==[] then []
else tail(dropWhile (/=' ') d)
测试 - >使用尾递归
beforep - >得到一切直到第一空间
afterp - >获得空间后的一切
有什么想法吗?如果您对此问题有任何其他解决方案,那将有所帮助。谢谢
答案 0 :(得分:5)
你差不多了。如果我尝试按原样运行您的代码,我会得到:
test.hs:2:23:
Couldn't match expected type `Char' with actual type `String'
Expected type: String
Actual type: [String]
In the return type of a call of `test'
In the first argument of `(:)', namely `(test (afterp d))'
所以检查第2行:
test d = beforep d : (test (afterp d)) : []
-- ^
-- This is the problem -----------------|
cons运算符的类型是:
(:) :: a -> [a] -> [a]
您的test
函数已经返回[String]
,您不想尝试将其置于空列表中。这意味着返回类型为[[String]]
。
请改为尝试:
test d = beforep d : (test (afterp d))
在更改之后,它会编译,但是当您运行test "i need help"
时,您将获得无限列表:
["i","need","help","","","","","","","",""...
问题是您需要在test
中包含一个基本案例,当您将其传递给空列表时会停止。这是工作代码:
test :: String -> [String]
test [] = []
test d = beforep d : (test (afterp d))
beforep :: String -> String
beforep d = takeWhile (/=' ') d
afterp :: String -> String
afterp d = if (dropWhile (/=' ') d)==[] -- Slightly reformatted
then [] -- to improve readability,
else tail(dropWhile (/=' ') d) -- no real change.