当前我在执行：
```bash
while [ -d "/proc/$PID" ]; do
  sleep 1
done

<details>
<summary>英文:</summary>
Currently I do:
    while [ -d &quot;/proc/$PID&quot;  ]; do
      sleep 1
    done
Since I cannot use `wait` as it is a non-child process.
The problem is that `sleep` blocks any signal trap handlers (for example for SIGINT or SIGTERM) until it is not sleeping anymore. So each signal will be delayed by 0.5 seconds on average.
I tried to use `tail` instead to wait for the PID, but it also blocks the trap handler (unless you press CTRL-C from interactive shell). I tried using `pwait` / `pidwait` but same problem: they don&#39;t exit when a stop signal is received.
So is there a good way to do this, and still be able to handle stop signals?
</details>
# 答案1
**得分**: 2
The solution was simple. Instead of calling `wait` on the PID, I need to call `wait` on the PID of an executable that is waiting for said PID. This allows the signal traps to be called while waiting.
For example:
    pidwait -F &quot;$PIDFILE&quot; &amp; wait $!
or:
    tail --pid &quot;$PID&quot; -f /dev/null &amp; wait $!
or even the original loop can be fixed this way by waiting on sleep:
    while [ -d &quot;/proc/$PID&quot;  ]; do
      sleep 1 &amp; wait $!
    done
This was a bit counter-intuitive, since you are waiting on something that is already waiting, but in hindsight it makes sense.
<details>
<summary>英文:</summary>
The solution was simple. Instead of calling `wait` on the PID, I need to call `wait` on the PID of an executable that is waiting for said PID. This allow the signal traps to be called while waiting.
For example:
    pidwait -F &quot;$PIDFILE&quot; &amp; wait $!
or:
    tail --pid &quot;$PID&quot; -f /dev/null &amp; wait $!
or even the original loop can be fixed this way by waiting on sleep:
    while [ -d &quot;/proc/$PID&quot;  ]; do
      sleep 1 &amp; wait $!
    done
This was a bit counter-intuitive, since you are waiting on something that is already waiting, but in hindsight it makes sense.
</details>