我想写函数并将结果放到字符串中.
我想要的功能:
read' :: FilePath -> String
我用:
:t readFile
readFile :: FilePath -> IO String
我做:
read' :: IO ()
read' = do
str <- readFile "/home/shk/workspace/src/test.txt"
putStrLn str
我想问str是不是字符串？
我们知道:
:t putStrLn
putStrLn :: String -> IO ()
那我为什么不能:
read' :: String
read' = do
str <- readFile "/home/shk/workspace/lxmpp/src/test.txt"
str
我得到错误:
Couldn't match expected type `[t0]' with actual type `IO String'
In the return type of a call of `readFile'
In a stmt of a 'do' expression:
str …checkstring :: [String] -> Int -> [String]
checkstring p n = do z <- doesFileExist (p !! n)
if z
then p
else error $ "'" ++ (p !! n) ++ "' file path does not exist"
它通过查看"n"检查字符串中的元素(因此,如果n = 2,它将检查列表中的第二个字符串),然后查看它是否存在.如果它确实存在,它将返回原始字符串列表,如果不存在则会出错.为什么这样做？:
Couldn't match expected type `[t0]' with actual type `IO Bool'
In the return type of a call of `doesFileExist'
In a stmt of a 'do' expression: z <- doesFileExist (p !! n)
我正在尝试编写一个带有IO Bool的函数,并根据这是什么做的,但我无法弄清楚如何评估IO Bool.我试着说do cond,do {cond==True}但得到了错误Couldn't match expected type 'Bool' against inferred type 'a b'.有人可以指导我找到解决方案吗？