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Function 如何使用bind和IO read Int重写'do'块？

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Function 如何使用bind和IO read Int重写'do'块？,function,haskell,io,monads,do-notation,Function,Haskell,Io,Monads,Do Notation,因此，我想重写给定的prog函数，使用>>=绑定而不是do和您更改代码太多，例如从IO Int中删除Int，并在错误的点插入lambda 像这样的方法应该会奏效： prog = putStrLn "Hello there! How old are you?" >> (readLn :: IO Int) >>= \age -> let agedays = show $ age * 365 in putStrLn $ "So you are at

因此，我想重写给定的
prog
函数，使用
>
>=绑定而不是
do
和
您更改代码太多，例如从
IO Int
中删除
Int
，并在错误的点插入lambda

像这样的方法应该会奏效：


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)



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 )


由于您已经指定了
prog:：IO Int
，这意味着
返回agedays:：IO Int
和
agedays:：Int

然后，在
age*365
中
*
的两个操作数必须是相同的类型，特别是
agedays
的操作数，因为这里有
agedays=age*365
。因此，它已经是
age:：Int
。
它不应该是
（readLn:：IO Int）>=\age->…
？

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)



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 )