bash 脚本中的 $$ 与子外壳中的 $$
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$$ in a script vs $$ in a subshell
提问by Ankur Agarwal
$$gives process id of the script process when used in a script, like this:
$$在脚本中使用时给出脚本进程的进程 ID，如下所示：
Example 1
示例 1
#!/bin/bash
# processid.sh
# print process ids
ps -o cmd,pid,ppid
echo "The value of $$ is $$"
$ ./processid.sh
CMD PID PPID
bash 15073 4657
/bin/bash ./processid.sh 15326 15073
ps -o cmd,pid,ppid 15327 15326
The value of $$ is 15326
Observe the pid given by $$and psis 15326
观察由$$和给出的 pidps是15326
My shell prompt is pid 15073
我的 shell 提示是 pid 15073
But in a subshell, $$gives pid of parent shell (which is 15073)
但是在子shell中，$$给出父shell的pid（15073）
Example 2
示例 2
$ ( ps -o cmd,pid,ppid ; echo $$ )
CMD PID PPID
bash 15073 4657
bash 15340 15073
ps -o cmd,pid,ppid 15341 15340
15073
Here subshell is pid 15340
这里的子shell是pid 15340
Question: Why so? Isn't the script also running in a subshell? What's the difference between the subshell in example 2 and the shell in which the script runs in example 1?
问题：为什么会这样？脚本不是也在子shell中运行吗？示例 2 中的子 shell 与示例 1 中运行脚本的 shell 有什么区别？
回答by flolo
I tried and escaping (to pass the $$ to the subshell) does not work as the subshell inherits the $$ value from the parent bash. The solution to this is to use $BASHPID.
我尝试转义（将 $$ 传递给子外壳）不起作用，因为子外壳从父 bash 继承了 $$ 值。解决方案是使用 $BASHPID。
(echo $$; echo $BASHPID)
prints the PID from the parent shell and from the subshell.
从父 shell 和子 shell 打印 PID。
回答by Tony Delroy
From the bash manpage:
从 bash 联机帮助页：
$ Expands to the process ID of the shell. In a () subshell, it
expands to the process ID of the current shell, not the sub-
shell.
回答by Ignacio Vazquez-Abrams
The replacement takes place in the parent shell; the subshell hasn't been started by the time the substitution takes place.
替换发生在父 shell 中；替换发生时子外壳尚未启动。
回答by drizzt
A more portable way, linux-only, but also compatible with dash:
一种更便携的方式，仅限 linux，但也与 dash 兼容：
read -r my_pid _ < /proc/self/stat
echo $my_pid
