Bash – Why is subshell created by background control operator (&) not displayed under pstree

bashjob-controljobssubshell

I understand that when I run exit it terminates my current shell because exit command run in the same shell. I also understand that when I run exit & then original shell will not terminate because & ensures that the command is run in sub-shell resulting that exit will terminate this sub-shell and return back to original shell. But what I do not understand is why commands with and without & looks exactly the same under pstree, in this case sleep 10 and sleep 10 &. 4669 is the PID of bash under which first sleep 10 and then sleep 10 & were issued and following output was obtained from another shell instance during this time:

# version without &
$ pstree 4669
bash(4669)───sleep(6345)

# version with &
$ pstree 4669
bash(4669)───sleep(6364)

Should't the version with & contain one more spawned sub-shell (e.g. in this case with PID 5555), like this one?

bash(4669)───bash(5555)───sleep(6364)

PS: Following code was omitted from output of pstree beginning for better readability:

systemd(1)───slim(1009)───ck-launch-sessi(1370)───openbox(1551)───/usr/bin/termin(4510)───bash(4518)───screen(4667)───screen(4668)───

Best Answer

Until I started answering this question, I hadn’t realised that using the & control operator to run a job in the background starts a subshell. Subshells are created when commands are wrapped in parentheses or form part of a pipeline (each command in a pipeline is executed in its own subshell).

The Lists of Commands section of the Bash manual (thanks jimmij) states:

If a command is terminated by the control operator ‘&’, the shell executes the command asynchronously in a subshell. This is known as executing the command in the background. The shell does not wait for the command to finish, and the return status is 0 (true).

As I understand it, when you run sleep 10 & the shell forks to create a new child process (a copy of itself) and then immediately execs to replace this child process with code from the external command (sleep). This is similar to what happens when a command is run as normal (in the foreground). See the Fork–exec Wikipedia article for a short overview of this mechanism.

I couldn’t understand why Bash would run backgrounded commands in a subshell but it makes sense if you also want to be able to run shell builtins such as exit or echo to be run in the background (not just external commands).

When it’s a shell builtin that’s being run in the background, the fork happens (resulting in a subshell) without an exec call to replace itself with an external command. Running the following commands shows that when the echo command is wrapped in curly braces and run in the background (with the &), a subshell is indeed created:

$ { echo $BASH_SUBSHELL $BASHPID; }
0 21516
$ { echo $BASH_SUBSHELL $BASHPID; } &
[1] 22064
$ 1 22064

In the above example, I wrapped the echo command in curly braces to avoid BASH_SUBSHELL being expanded by the current shell; curly braces are used to group commands together without using a subshell. The second version of the command (ending with the & control operator) clearly demonstrates that terminating the command with the ampersand has resulted in a subshell (with a new PID) being created to execute the echo builtin. (I’m probably simplifying the shell’s behaviour here. See mikeserv’s comment.)

I would never have thought of running exit & and had I not read your question, I would have expected the current shell to quit. Knowing now that such commands are run in a subshell, your explanation that it’s the subshell which exits makes sense.

“Why is subshell created by background control operator (&) not displayed under pstree”

As mentioned above, when you run sleep 10 &, Bash forks itself to create the subshell but since sleep is an external command, it calls the exec() system call which immediately replaces the Bash code and data in the child process with a running copy of the sleep program. By the time you run pstree, the exec call will already have completed and the child process will now have the name “sleep”.

While away from my computer, I tried to think of a way of keeping the subshell running long enough for the subshell to be displayed by pstree. I figured we could run the command through the time builtin:

$ time sleep 11 &
[2] 4502
$ pstree -p 26793
bash(26793)─┬─bash(4502)───sleep(4503)
            └─pstree(4504)

Here, the Bash shell (26793) forks to create a subshell (4502) in order to execute the command in the background. This subshell runs its own time builtin command which, in turn, forks (to create a new process with PID 4503) and execs to run the external sleep command.

Using named pipes, jimmij came up with a clever way to keep the subshell created to run exit alive long enough for it to be displayed by pstree:

$ mkfifo file
$ exit <file &
[2] 6413
$ pstree -p 26793
bash(26793)─┬─bash(6413)
            └─pstree(6414)
$ echo > file
$ jobs
[2]-  Done    exit < file

Redirecting stdin from a named pipe is clever as it causes the subshell to block until it receives input from the named pipe. Later, redirecting the output of echo (without any arguments) writes a newline character to the named pipe which unblocks the subshell process which, in turn, runs the exit builtin command.

Similarly, for the sleep command:

$ mkfifo named_pipe
$ sleep 11 < named_pipe &
[1] 6600
$ pstree -p 26793
bash(26793)─┬─bash(6600)
            └─pstree(6603)

Here we see that the subshell created to run the command in the background has a PID of 6600. Next, we unblock the process by writing a newline character to the pipe:

$ echo > named_pipe

The subshell then execs to run the sleep command.

$ pstree -p 26793
bash(26793)─┬─pstree(6607)
            └─sleep(6600)

After the exec() call, we can see that the child process (6600) is now running the sleep program.

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