子shell的$$

深碍√TFBOYSˉ_ 2022-06-13 09:17 31阅读 0赞

下面是一段代码,我偶然发现有点问题。

  1. #!/bin/bash
  2. TestID()
  3. {
  4. echo "in function : $$"
  5. }
  6. TestID &
  7. echo "\$! : $! "
  8. echo "\$\$ : $$ "
  9. sleep 1

[firefox@fire ShellPractice]$ ./stopped.sh
$! : 15021
$$ : 15020
in function : 15020

疑问就是在这里,$!代表是最后一个后台程序的ID,是当前进程ID,而&出来的应该是一个子shell,但是它的ID(在子shell里这时候是打印出来的)和父shell是“一样”的。我去论坛上问了下,然后查了些资料。下面是一些简单的介绍:

shell什么情况下会产生子进程

1:&,提交后台作业

If a command is terminated by the control operator `&’, the shell executes the command asynchronously in a subshell.

2:管道

Each command in a pipeline is executed in its own subshell

3:括号命令列表

()操作符

Placing a list of commands between parentheses causes a subshell environment to be created

4:执行外部脚本、程序:

When Bash finds such a file while searching the `$PATH’ for a command, it spawns a subshell to execute it. In other words, executing filename ARGUMENTS is equivalent to executing bash filename ARGUMENTS

再看看 man bash 里面对于$的解释

“$” 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.

那么这里的current shell怎么解释呢?字面意思是“当前的shell”。这里有有一篇博客讲子shell,我觉得讲的很好 http://blog.csdn.net/sosodream/article/details/5683515 。用 ./xxx.sh 执行脚本的时候也会生成子shell,原来的shell会有一个 ID,这里假设叫ID1,然后进入脚本,这时脚本里的就是current shell的ID(由ID1生出的子shell的ID)之后再生出其余的子shell的都是current shell的ID,也就是由ID1生出来的ID,注意:子shell的ID可不是$$。子shell有它自己的ID。至于怎么获取,一般都用$!来获取。

再来看看下面的脚本

  1. #!/bin/bash
  2. echo "\$\$ outside of subshell = $$"
  3. echo "\$BASH_SUBSHELL outside of subshell = $BASH_SUBSHELL"
  4. echo "\$BASHPID outside of subshell = $BASHPID"
  5. echo "-------------------"
  6. (
  7. echo "\$\$ inside of subshell = $$"
  8. echo "\$BASH_SUBSHELL inside of subshell = $BASH_SUBSHELL"
  9. echo "\$BASHPID inside of subshell = $BASHPID"
  10. )
  11. echo "-------------------"

$BASHPID就是指当前shell。

现在再来看看 man bash 中“$”的解释

[root@fire ShellPractice]# echo $$
2618

如果你在子shell里面执行这条命令,其实是先替换为2618,然后再进行echo,一直都指的的是current shell。

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