主页 / unix / 问题 / 429973
Accepted
corsel
Asked: 2018-03-14 06:17:32 +0800 CST

无法让 bash 集群工作

  • 772

我从我所看到的关于 bashflock函数的内容中派生了一个示例 bash 脚本。我愿意：

func()
{
    42>/home/foo
    flock -e 42 || exit 1
    echo "hello world"
    sleep 5
}

然后我连续运行func&，每个都hello world立即打印，而我希望第一个打印消息，其余的退出。我在这里想念什么？

bash file-descriptors

1 个回答

  1. Best Answer

    考虑这个示例，基于flock手册页中的示例：

    #!/bin/bash

func() {
    echo "$$ trying to acquire lock"
    (
        flock -e 42
        echo "lock acquired by $$"
        sleep 10
    ) 42> /tmp/mylock
    echo "lock released by $$"
}

func

    现在，如果我运行该脚本一次：

    $ bash ex.sh
22241 trying to acquire lock
lock acquired by 22241
lock released by 22241

    如果我在这个脚本的 10 秒睡眠窗口内运行两个实例，第一个在后台运行，可能的事件序列是：

    $ bash ex.sh& bash ex.sh
[1] 24518
24519 trying to acquire lock
24518 trying to acquire lock
lock acquired by 24519
lock released by 24519
lock acquired by 24518
$
lock released by 24518

    在此示例中，第二个进程赢得了比赛并首先获得了锁。然后它释放锁并允许第一个（后台）进程获取，然后释放锁。

    我可以通过在启动它们之间引入延迟来提高第一个进程赢得比赛的机会：

    $ bash ex.sh& sleep 1; bash ex.sh
[1] 30158
30158 trying to acquire lock
lock acquired by 30158
30179 trying to acquire lock
lock released by 30158
lock acquired by 30179
lock released by 30179
    • 3

相关问题