如何使用綁定和IO讀取Int重寫`do`塊？

Question

因此，我想使用>> / >>=綁定而不是do和<-重prog定的prog函數：

prog :: IO Int
     prog =
       do putStrLn "Hello there! How old are you?"
       age <- (readLn :: IO Int)
       let agedays = show $ age * 365
       putStrLn $ "So you are at least than " ++ agedays ++ " days old."
       return (read agedays)

重寫更簡單的功能對我來說不是問題，但是readLn :: IO Int讓我頭疼。

我的建議是：

prog :: IO Int
prog =
     putStrLn "Hello there!How old are you?" >>
     readLn::IO >>=
     let agedays = \age -> show $ age * 365 >>
     putStrLn $ "So you are at least than " ++ agedays ++ " days old."

但是，這是行不通的，因為將readLn :: IO綁定到下一個匿名函數\\age 。 有什么幫助嗎？

Answer 1

您正在改變的代碼太多，例如去除Int從IO Int ，並插入錯誤的點lambda表達式。

這樣的事情應該起作用：

prog =
   putStrLn "Hello there! How old are you?" >>
   (readLn :: IO Int) >>= \age ->
   let agedays = show $ age * 365
   in putStrLn $ "So you are at least than " ++ agedays ++ " days old." >>
   return (read agedays)

Answer 2

您可以讓類型推斷為您完成工作，

prog :: IO Int
prog =
     putStrLn "Hello there! How old are you?" >>
     readLn >>= (\ age ->
     let agedays = age * 365 in
       putStrLn ("So you are at least " ++ show agedays ++ " days old.") >>
       return agedays )

由於您已經指定了prog :: IO Int ，因此意味着return agedays :: IO Int和agedays :: Int 。

然后，這兩個操作數*在age * 365必須是同一類型的，具體而言，是的agedays ，因為我們有agedays = age * 365那里。 因此，它已經遵循該age :: Int了。