0

In my Bash shell the $$ variable expands to the process ID of the shell.

So why does ps print its own name when I execute this?

$ bash -c 'ps -p $$ -o comm='
ps

When I follow it with a no-op command, ps does print the name of the shell.

$ bash -c 'ps -p $$ -o comm= && true'
bash

Versions:

  • GNU bash, version 5.0.17(1)-release (x86_64-pc-linux-gnu)
  • ps from procps-ng 3.3.16
Iain Samuel McLean Elder
  • 267
  • 2
  • 14
  • [Why bash does not spawn a subshell for simple comamands?](https://unix.stackexchange.com/a/466519) is another way of asking the same question. [Sergiy Kolodyazhnyy](https://unix.stackexchange.com/a/466519) shows that the first example is a performance optimization on a simple command. I expected Bash to fork a new process, but instead it starts a process with the same pid and replaces itself. The second example is a compound command. Bash can't replace itself because it needs to supervise all the commands in the compound, so it forks each subprocess with a new pid and retains its own. – Iain Samuel McLean Elder Aug 31 '23 at 12:47
  • In particular, Bash needs to keep itself alive until `ps` exits so it can then run `true` if needed. It wouldn't need to keep itself alive until `true` exits, though, but Bash only optimizes the case of a single command, not any tail-end command. However, ksh93 optimizes that too, e.g. `ksh -c 'ps -p $$ -o comm=; ps -p $$ -o comm='` prints `ksh`, `ps` – ilkkachu Aug 31 '23 at 17:11

0 Answers0