用信号陷阱中断 bash 中的睡眠
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Interrupt sleep in bash with a signal trap
提问by John Morris
I'm trying to catch the SIGUSR1signal in a bash script that is sleeping via the sleepcommand:
我正在尝试SIGUSR1通过以下sleep命令在休眠的 bash 脚本中捕获信号：
#!/bin/bash
trap 'echo "Caught SIGUSR1"' SIGUSR1
echo "Sleeping. Pid=$$"
while :
do
sleep 10
echo "Sleep over"
done
The signal trap works, but the message being echoed is not displayed until the sleep 10has finished.
It appears the bash signal handling waits until the current command finished before processing the signal.
信号陷阱有效，但在sleep 10完成之前不会显示正在回显的消息。
bash 信号处理似乎在处理信号之前等待当前命令完成。
Is there a way to have it interrupt the running sleepcommand as soon as it gets the signal, the same way a C program would interrupt the libc sleep()function?
有没有办法sleep让它在收到信号后立即中断正在运行的命令，就像 C 程序中断 libcsleep()函数一样？
回答by Hans Klünder
#!/bin/bash
trap 'echo "Caught SIGUSR1"' SIGUSR1
echo "Sleeping. Pid=$$"
while :
do
sleep 10 &
wait $!
echo "Sleep over"
done
回答by Echoes_86
Just a point about the waitafter the sleepbecause I've just made this error in my script:
只是关于wait之后的一点，sleep因为我刚刚在我的脚本中犯了这个错误：
You should use
wait $!instead ofwaitif, inside your script, you already launched other processes in background
您应该在脚本中使用
wait $!而不是waitif，您已经在后台启动了其他进程
For example, the waitinside the next snippet of code will waitfor the termination of both process_1and sleep 10:
例如，wait下一个代码片断里面会等待两个终止process_1和sleep 10：
process_1 &
...
sleep 10 &
wait
If you use, instead of wait, wait $!your script will wait only for sleep 10, because $!means PID of last backgrounded process.
如果您使用，而不是wait，wait $!您的脚本将只等待sleep 10，因为$!意味着最后一个后台进程的 PID。
回答by dirk
Remark that
请注意
sleep infinity &
wait
puts the sleep in background, and stops the wait with the signal. This leaves an infinite sleep behind on every signal !
将睡眠置于后台，并用信号停止等待。这会在每个信号上留下无限睡眠！
Replace the sleep and wait with
替换睡眠并等待
read
and you will be fine.
你会没事的。
