如何使用带有 IO read Int 的绑定重写 `do` 块？

Question

所以，我想使用 >>/>>= 绑定而不是 do 和 <- 来重写给定的 prog 函数：

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)

重写更简单的功能对我来说不是问题，但是readLn :: IO Int让我很头疼...

我的建议是：

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."

但是这不起作用，因为将 readLn :: IO 绑定到下一个匿名函数 \age 时出现问题。有帮助吗？

Answer 1

您对代码的更改过多，例如从 IO Int 中删除 Int，并在错误的位置插入 lambda。

像这样的东西应该可以工作：

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

您可以让类型推断为您完成工作，

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 )

既然你已经指定了prog :: IO Int，那就意味着return agedays :: IO Int，而agedays :: Int。

那么，age * 365 中 * 的两个操作数必须属于同一类型，具体来说，就是 agedays 的类型，因为那里有 agedays = age * 365。因此它遵循 age :: Int 已经。

如何使用带有 IO read Int 的绑定重写 `do` 块？

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