linux - python编写守护进程进程

Question

最近在看shadowsocks源代码,关于作者写的启动守护进程的这一块不是很理解:

def daemon_start(pid_file, log_file):

    def handle_exit(signum, _):
        if signum == signal.SIGTERM:
            sys.exit(0)
        sys.exit(1)

    signal.signal(signal.SIGINT, handle_exit)
    signal.signal(signal.SIGTERM, handle_exit)

    # fork only once because we are sure parent will exit
    pid = os.fork()
    assert pid != -1

    if pid > 0:
        # 这里为什么不是让父进程直接退出,而是等着子进程把自己杀死呢?
        time.sleep(5)
        sys.exit(0)

    # child signals its parent to exit
    ppid = os.getppid()
    pid = os.getpid()
    if write_pid_file(pid_file, pid) != 0:
        os.kill(ppid, signal.SIGINT)
        sys.exit(1)

    os.setsid()
    signal.signal(signal.SIGHUP, signal.SIG_IGN)

    print('started')
    os.kill(ppid, signal.SIGTERM)

我在看APUE关于守护进程编写那一块的时候,书上说的是父进程fork子进程后退出,不知道这里为什么不是直接退出而是等着子进程把自己杀死?

Answer 1

Here, the parent process registers the signal processing function

    signal.signal(signal.SIGINT, handle_exit)
    signal.signal(signal.SIGTERM, handle_exit)

In order to give the parent process a chance to process the signal sent by the child process, it sleeps for 5 seconds.