As I understand pipes and commands, bash takes each command, spawns a process for each one and connects stdout of the previous one with the stdin of the next one.

For example, in "ls -lsa | grep feb", bash will create two processes, and connect the output of "ls -lsa" to the input of "grep feb".

When you execute a background command like "sleep 30 &" in bash, you get the pid of the background process running your command. Surprisingly for me, when I wrote "ls -lsa | grep feb &" bash returned only one PID.

How should this be interpreted? A process runs both "ls -lsa" and "grep feb"? Several process are created but I only get the pid of one of them?

2 Answers 2


When you run a job in the background, bash prints the process ID of its subprocess, the one that runs the command in that job. If that job happens to create more subprocesses, that's none of the parent shell's business.

When the background job is a pipeline (i.e. the command is of the form something1 | something2 &, and not e.g. { something1 | something2; } &), there's an optimization which is strongly suggested by POSIX and performed by most shells including bash: each of the elements of the pipeline are executed directly as subprocesses of the original shell. What POSIX mandates is that the variable $! is set to the last command in the pipeline in this case. In most shells, that last command is a subprocess of the original process, and so are the other commands in the pipeline.

When you run ls -lsa | grep feb, there are three processes involved: the one that runs the left-hand side of the pipe (a subshell that finishes setting up the pipe then executes ls), the one that runs the right-hand side of the pipe (a subshell that finishes setting up the pipe then executes grep), and the original process that waits for the pipe to finish.

You can watch what happens by tracing the processes:

$ strace -f -e clone,wait4,pipe,execve,setpgid bash --norc
execve("/usr/local/bin/bash", ["bash", "--norc"], [/* 82 vars */]) = 0
setpgid(0, 24084)                       = 0
bash-4.3$ sleep 10 | sleep 20 &

Note how the second sleep is reported and stored as $!, but the process group ID is the first sleep. Dash has the same oddity, ksh and mksh don't.


Spawns 2 processes. The & displays the PID of the second process. Example below.

$ echo $$
$ sleep 100 | sleep 200 &
[1] 13405
$ ps -ef|grep 13358
ec2-user 13358 13357  0 19:02 pts/0    00:00:00 -bash
ec2-user 13404 13358  0 19:04 pts/0    00:00:00 sleep 100
ec2-user 13405 13358  0 19:04 pts/0    00:00:00 sleep 200
ec2-user 13406 13358  0 19:04 pts/0    00:00:00 ps -ef
ec2-user 13407 13358  0 19:04 pts/0    00:00:00 grep --color=auto 13358
  • 1
    Tested it with 5 sleeps, and always returns the PID of the last one. Thanks.
    – Raul
    Nov 27, 2016 at 19:21

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