如何编写返回Haskell

我希望我的showStackHead函数获取一个堆栈打印头部并返回其余部分，这是我的代码 --代码

---

当我运行这段代码时，编译器告诉我第二个返回有一个解析错误，那么编写此函数的正确方法是什么？

将“返回”缩进到与“putStrLn”相同的深度，如下所示：

showStackHead xs
   | xs == []   = return []
   | otherwise  = do putStrLn ("result:" ++ (head xs))
                     return (tail xs)

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另外，通过使用模式匹配，您的showStackHead可能会更干净。允许您放弃头部和尾部的防护比较：

#! /usr/bin/env runhaskell


showStackHead []     = return []
showStackHead (x:xs) = do
   putStrLn $ "result: " ++ [x]
   return xs


main :: IO ()
main = do
   let s1 = []
   r1 <- showStackHead s1
   putStrLn $ "returned: " ++ r1

   putStrLn "---"

   let s2 = "foo"
   r2 <- showStackHead s2
   putStrLn $ "returned: " ++ r2

   putStrLn "---"

   let s3 = "q"
   r3 <- showStackHead s3
   putStrLn $ "returned: " ++ r3

#/usr/bin/env runhaskell
showStackHead[]=返回[]
showStackHead（x:xs）=do
putStrLn$“结果：”++[x]
返回xs
main:：IO（）
main=do
设s1=[]
r1
#! /usr/bin/env runhaskell


showStackHead []     = return []
showStackHead (x:xs) = do
   putStrLn $ "result: " ++ [x]
   return xs


main :: IO ()
main = do
   let s1 = []
   r1 <- showStackHead s1
   putStrLn $ "returned: " ++ r1

   putStrLn "---"

   let s2 = "foo"
   r2 <- showStackHead s2
   putStrLn $ "returned: " ++ r2

   putStrLn "---"

   let s3 = "q"
   r3 <- showStackHead s3
   putStrLn $ "returned: " ++ r3