如何在程序包“loops”中使用`break_`？ [英] How do I use `break_` in the package "loops"?

问题描述

查看软件包 loops ，我发现它很有趣，可能会有用。但是，有一部分关于我不明白的包：我该如何使用 break _ ？

我有一个函数 get':: IO（Maybe Int），每次调用都会返回一个从文件中读取的数字，并返回 Nothing 如果EOF已达到。我试图构造一个简单的循环，在这里我打印每个数字并在EOF中断开。

现在，我知道无限循环，我可以使用永远：

 导入Control.Monad.Loop作为ML 
导入Control.Monad as M 
 
 main = do 
 M.sequence。 loop $ do 
 ML.forever 
 return $ do 
 mx<  -  get'
 case mx of 
 Nothing  - > ??? 
只需x  - >打印x 
  

但是我在哪里放置 break _ ？它是一个 LoopT IO Int ，所以我只能将它放在 LoopT monad中，但它不是它应该被称为中间interation，而不是定义循环时？这真令我困惑。

解决方案

LoopT 是一个monad transfomer，所以您需要 liftIO 您的 print x 语句。

以下是一些用法示例：

  import Control.Monad 
 import Control.Monad.Trans（liftIO）
导入Control.Monad.Loop作为ML 
 
  - 无限打印循环
 foo :: LoopT IO（）
 foo = do 
x<  -  ML。迭代0（+1）
 liftIO $ print x 
 
 run_foo = ML.exec_ foo 
 
  - 打印1,3,5,7,9 
 bar :: IO（）
 bar = ML.exec_ $ do 
x<  -  ML.iterate 1（+2）
 if x< 10 
然后liftIO $ print x 
 else break_ 
  

更新

如果你想在另一个monad中执行一个无限循环，我只需要使用永远 code> Control.Monad -

  import Control.Monad 
 import Control.Monad.State 
 
 myloop = forever $ do 
x<  -  get 
 liftIO $ print x 
 put（x + 1）
 
 main = runStateT myloop 10  - 启动循环为10 
  

更新2

另一个使用monadic条件的例子：

  findFavoriteNumber = ML .exec_ $ do 
x<  -  ML.iterate 1（+1）
 yn<  -  liftIO $ do putStr $is++ show x ++your favorite number？
 getLine 
 if yn ==yes
 then break_ 
 else return（）
  

当然，这个循环不会返回最喜欢的数字 - 它只是保持不变直到用户回答是。

Looking into the package loops, I found it very interesting and could be useful. However, there's one part about the package that I don't understand: how am I supposed to use break_?

Let's say I have a function get' :: IO (Maybe Int), with which each call will return a number read from a file, and returns Nothing if the EOF is reached. I'm trying to construct a simple loop where I print each number and break upon EOF.

Now, I know to loop indefinitely I can use forever:

import Control.Monad.Loop as ML
import Control.Monad as M

main = do
    M.sequence . loop $ do
        ML.forever
        return $ do
            mx <- get'
            case mx of
                Nothing -> ???
                Just x  -> print x

But where do I put my break_? It is a LoopT IO Int, so I can only put it in LoopT monad, but isn't it supposed to be called mid interation, instead of when defining loops? This really confuses me.

解决方案

LoopT is a monad transfomer, so you will need to liftIO your print x statement.

Here are some usage examples:

import Control.Monad 
import Control.Monad.Trans (liftIO)
import Control.Monad.Loop as ML

-- infinite printing loop
foo :: LoopT IO ()
foo = do
  x <- ML.iterate 0 (+1) 
  liftIO $ print x

run_foo = ML.exec_ foo

-- prints 1, 3, 5, 7, 9
bar :: IO () 
bar = ML.exec_ $ do
  x <- ML.iterate 1 (+2)
  if x < 10
    then liftIO $ print x
    else break_

Update

If you want to execute a infinite loop in another monad, I would just use forever from Control.Monad --

import Control.Monad
import Control.Monad.State

myloop = forever $ do
  x <- get
  liftIO $ print x
  put (x+1)

main = runStateT myloop 10 -- start loop at 10

Update 2

Another example using a monadic condition:

findFavoriteNumber = ML.exec_ $ do
  x <- ML.iterate 1 (+1)
  yn <- liftIO $ do putStr $ "is " ++ show x ++ " your favorite number? "
                    getLine
  if yn == "yes"
    then break_
    else return ()

Of course, this loop doesn't return the favorite number - it just keeps asking until the user responds with "yes".

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