Debian Buster에 언바운드를 설치했습니다. 이제 언바운드가 각 프로토콜을 2번씩 수신하는 이유가 궁금합니다.
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
누구든지 내가 이것을 이해하도록 도와줄 수 있나요?
답변1
언바운드는SO_REUSEPORT여러/프로세스 스레드가 동일한 수신 포트를 공유하도록 허용하는 옵션입니다.
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.