captain captain - 3 months ago 16
Linux Question

Difference between bash pid and $$

I'm a bash scripting beginner, and I have a "homework" to do. I figured most of the stuff out but there is a part which says that I have to echo the pid of the parent bash and the pid of the two subshells that I will be running. So I looked online and found this (The Linux documentation project):

#!/bin/bash4

echo "\$\$ outside of subshell = $$" # 9602
echo "\$BASH_SUBSHELL outside of subshell = $BASH_SUBSHELL" # 0
echo "\$BASHPID outside of subshell = $BASHPID" # 9602

echo

( echo "\$\$ inside of subshell = $$" # 9602
echo "\$BASH_SUBSHELL inside of subshell = $BASH_SUBSHELL" # 1
echo "\$BASHPID inside of subshell = $BASHPID" ) # 9603
# Note that $$ returns PID of parent process.


So here are my questions:

1) What does the first echo print? Is this the pid of the parent bash?

2) Why does the 2nd echo print out 0?

3) Is $BASH_SUBSHELL a command or a variable?

4) I'm doing everything on a mac, I will try all of this on a Linux machine in some days but
whenever I run this script
$BASHPID
doesn't return anything, I just get a new line. Is this because I'm running this on a mac and
$BASHPID
doesn't work on a mac?

imm imm
Answer

Looking at documentation on this, it looks like:

  1. $$ means the process ID that the script file is running under. For any given script, when it is run, it will have only one "main" process ID. Regardless of how many subshells you invoke, $$ will always return the first process ID associated with the script. BASHPID will show you the process ID of the current instance of bash, so in a subshell it will be different than the "top level" bash which may have invoked it.
  2. BASH_SUBSHELL indicates the "subshell level" you're in. If you're not in any subshell level, your level is zero. If you start a subshell within your main program, that subshell level is 1. If you start a subshell within that subshell, the level would be 2, and so on.
  3. BASH_SUBSHELL is a variable.
  4. Maybe BASHPID isn't supported by the version of bash you have? I doubt it's a "Mac" problem.
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