python守护进程-为什么这个函数会两次杀死父进程?

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:
        # parent waits for its child
        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)

    sys.stdin.close()
    try:
        freopen(log_file, 'a', sys.stdout)
        freopen(log_file, 'a', sys.stderr)
    except IOError as e:
        shell.print_exception(e)
        sys.exit(1)

此守护进程不使用双叉。上面写着“叉子只有一次,因为我们确信父母会退出”。父级调用sys.exit(0)退出。但是子级调用os.kill(ppid,signal.SIGTERM)退出父级。