了解 Unbound 如何监听 53 端口

了解 Unbound 如何监听 53 端口

我在 Debian Buster 上安装了 unbound。现在我想知道为什么 unbound 在每个协议上监听 2 次。

netstat -tulpn

tcp        0      0 0.0.0.0:53              0.0.0.0:*               LISTEN      1150/unbound        
tcp        0      0 0.0.0.0:53              0.0.0.0:*               LISTEN      1150/unbound        
tcp        0      0 0.0.0.0:22              0.0.0.0:*               LISTEN      605/sshd: /usr/sbin 
tcp        0      0 127.0.0.1:8953          0.0.0.0:*               LISTEN      1150/unbound        
tcp6       0      0 :::22                   :::*                    LISTEN      605/sshd: /usr/sbin 
udp        0      0 0.0.0.0:53              0.0.0.0:*                           1150/unbound        
udp        0      0 0.0.0.0:53              0.0.0.0:*                           1150/unbound   

有人能帮助我理解这一点吗?

答案1

Unbound 使用重用端口允许多个进程线程共享同一个监听端口的选项。

SO_REUSEPORT(since Linux 3.9)

          Permits multiple AF_INET or AF_INET6 sockets to be bound
          to an identical socket address.  This option must be set
          on each socket (including the first socket) prior to
          calling bind(2) on the socket.  To prevent port hijacking,
          all of the processes binding to the same address must have
          the same effective UID.  This option can be employed with
          both TCP and UDP sockets.

          For TCP sockets, this option allows accept(2) load
          distribution in a multi-threaded server to be improved by
          using a distinct listener socket for each thread.  This
          provides improved load distribution as compared to
          traditional techniques such using a single accept(2)ing
          thread that distributes connections, or having multiple
          threads that compete to accept(2) from the same socket.

          For UDP sockets, the use of this option can provide better
          distribution of incoming datagrams to multiple processes
          (or threads) as compared to the traditional technique of
          having multiple processes compete to receive datagrams on
          the same socket.

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