用于 http 服务器的 haskell acceptFork

Question

我必须在 haskell 中创建一个 http 服务器。为此，我必须定义一个 acceptFork 函数来处理传入的 TCP 连接，它具有以下类型：

acceptFork :: Socket -> (Handle -> IO ()) -> IO ()

它的工作方式如下：

• 使用accept函数接受传入的连接，该连接返回一个Handle
• 使用 hSetBuffering 函数禁用缓冲（将值设置为 NoBuffering）
• 使用 hSetNewlineMode 函数设置新行字符（使用参数 (NewlineMode CRLF CRLF)）
• 启动一个新线程，对句柄值运行输入操作
• 递归地等待下一个连接

到目前为止，我想出的是：

import qualified Network.Socket as NS

acceptFork :: Socket -> (Handle -> IO ()) -> IO ()
acceptFork lsock k = do
    (csock,_) <- NS.accept lsock
    hSetBuffering lsock NoBuffering
    hSetNewlineMode lsock (NewlineMode CRLF CRLF)
    forkIO (k csock) 

问题是它甚至无法编译。有人可以解释我做错了什么吗？谢谢！

这是我得到的错误：

bead2.hs:31:23:
Couldn't match expected type `(Handle -> BufferMode -> IO ())
                              -> Socket
                              -> BufferMode
                              -> (Handle -> NewlineMode -> IO ())
                              -> Socket
                              -> NewlineMode
                              -> IO (Handle, t0)'
            with actual type `IO (Socket, NS.SockAddr)'
The function `NS.accept' is applied to 7 arguments,
but its type `Socket -> IO (Socket, NS.SockAddr)' has only one
In a stmt of a 'do' block:
  conn@(csock, _) <- NS.accept
                       lsock
                       hSetBuffering
                       lsock
                       NoBuffering
                       hSetNewlineMode
                       lsock
                       (NewlineMode CRLF CRLF)
In the expression:
  do { conn@(csock, _) <- NS.accept
                            lsock
                            hSetBuffering
                            lsock
                            NoBuffering
                            hSetNewlineMode
                            lsock
                            (NewlineMode CRLF CRLF);
       forkIO (k csock) }

bead2.hs:34:5:
Couldn't match type `ThreadId' with `()'
Expected type: IO ()
  Actual type: IO ThreadId
In a stmt of a 'do' block: forkIO (k csock)
In the expression:
  do { conn@(csock, _) <- NS.accept
                            lsock
                            hSetBuffering
                            lsock
                            NoBuffering
                            hSetNewlineMode
                            lsock
                            (NewlineMode CRLF CRLF);
       forkIO (k csock) }

Answer 1

好的，我复制了你的内容并添加了缺失的部分（对于网络 > 2.6 和

import qualified Network.Socket as NS
import System.IO (Handle, hSetBuffering, hSetNewlineMode, BufferMode(..), Newline(..), NewlineMode(..), IOMode(..))
import Control.Concurrent (forkIO, ThreadId)

acceptFork :: NS.Socket -> (Handle -> IO ()) -> IO ThreadId
acceptFork lsock k = do
    (csock,_) <- NS.accept lsock
    hsock <- NS.socketToHandle csock ReadWriteMode
    hSetBuffering hsock NoBuffering
    hSetNewlineMode hsock (NewlineMode CRLF CRLF)
    forkIO (k hsock)

基本上你有：

• 大量缺失的导入
• 需要通过NS.socketToHandle从socket获取句柄
• 注意forkIO（新线程的ID）的结果

编译如下：

 base >=4.8 && <4.9, network >=2.6 && <2.7

免责声明

我确实没有运行此代码，所以我不知道它是否有效 - 但它会编译（在我的系统上；））

没有ThreadId:

只需更改签名和最后一行：

acceptFork :: NS.Socket -> (Handle -> IO ()) -> IO ()
acceptFork lsock k = do
    -- ... same as above
    _ <- forkIO (k hsock)
    return ()

fork 并等待下一个连接：

只需递归调用acceptFork

acceptFork :: NS.Socket -> (Handle -> IO ()) -> IO ()
acceptFork lsock k = do
    (csock,_) <- NS.accept lsock
    hsock <- NS.socketToHandle csock ReadWriteMode
    hSetBuffering hsock NoBuffering
    hSetNewlineMode hsock (NewlineMode CRLF CRLF)
    _ <- forkIO (k hsock)
    acceptFork lsock k

或使用forever

...
import Control.Monad (forever)

acceptFork :: NS.Socket -> (Handle -> IO ()) -> IO ()
acceptFork lsock k = ...

acceptAll :: NS.Socket -> (Handle -> IO ()) -> IO ()
acceptAll sock = forever . acceptFork sock