Question

为什么下面的代码段会产生错误parse error on input ‘putStrLn’？

main = do line <- fmap reverse getLine
   putStrLn $ "You said " ++ line ++ " backwards!"
   putStrLn $ "Yes, you said " ++ line ++ " backwards!"

<interactive>:11:4: error: parse error on input ‘putStrLn’

此外，为什么以下代码段会产生错误parse error on input ‘let’？

main = do line <- getLine
  let line' = reverse line
  putStrLn $ "You said " ++ line' ++ " backwards!"
  putStrLn $ "Yes, you said " ++ line' ++ " backwards!"


 <interactive>:31:4: error: parse error on input ‘let’

Answer 1

两个片段都有相同的问题。如果将do块的第一个动作与do本身放在同一行，则仍必须缩进do块中的其余动作，直到第一个动作为止一。修复它的两种选择：

main = do line <- fmap reverse getLine
          putStrLn $ "You said " ++ line ++ " backwards!"
          putStrLn $ "Yes, you said " ++ line ++ " backwards!"

或

main = do
   line <- fmap reverse getLine
   putStrLn $ "You said " ++ line ++ " backwards!"
   putStrLn $ "Yes, you said " ++ line ++ " backwards!"

Answer 2

当在整个过程中使用显式分隔符时，它也适用：

main = do { line <- fmap reverse getLine ;
   putStrLn $ "You said " ++ line ++ " backwards!" ;
   putStrLn $ "Yes, you said " ++ line ++ " backwards!" }

main = do { line <- getLine ;
   let { line' = reverse line } ;                     -- NB let's { }s
   putStrLn $ "You said " ++ line' ++ " backwards!" ;
   putStrLn $ "Yes, you said " ++ line' ++ " backwards!" }

这不是好的缩进样式的替代品，而是对它的补充。

一个涉及putStrLn的谜

2 个答案: