How to suppress function output in Bash?

Question

I wrote a little helper to start an Android emulator which looks like this:

# START ANDROID EMULATION
android-sim () {
  if [[ -z "$1" ]]; then
    echo "No Android device supplied as argument!"
    return 1
  fi

  if [[ -d "$ANDROID_HOME" ]]; then
    cd "$ANDROID_HOME/emulator" &>/dev/null 
    ./emulator "@$1" &>/dev/null &
    cd - &>/dev/null 
  else
    echo "The variable ANDROID_HOME is not set to a correct directory!"
  fi
}

I tried to suppress all the generated output with &>/dev/null. For the emulator itself I also tried to put the execution in the background.

But when I run the function I still get the following output:

# before exeuction
[09:11:48] sandrowinkler:~ $ android-sim pixel-29
[1] 23217

# after execution, I closed the emulator via GUI
[09:11:55] sandrowinkler:~ $ 
[1]+  Done                    ./emulator "@$1" &> /dev/null  
(wd: /usr/local/share/android-sdk/emulator)
(wd now: ~)

According to this superuser post and this unix post I used the right way to silence the output. So what am I doing wrong here?

PS: I would also appreciate tips to improve my Bash code.

You can suppress the "Done" message by turning off [Job Control](https://www.gnu.org/software/bash/manual/bash.html#Job-Control) with `set +m`, but then you don't have job control in your interactive shell. — glenn jackman, Sep 02 '20 at 11:08

score 1 · Answer 1 · answered Sep 02 '20 at 07:30

1

You've done it correctly. That output is not from your script, but from the shell, displaying information about the background process (the PID, the done message and the working directory when in ends).

If you redirect the input externally, like this:

andoroid-sim foo > log.txt

and after execution you open log.txt, you'll see what actually your program outputed (and those messages won't be there, even if they appear in the console).

answered Sep 02 '20 at 07:30

Miguel

2,130
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Is there no way to suppress those messages as well? – winklerrr Sep 02 '20 at 08:00
1

See @NahuelFouilleul answer – Miguel Sep 02 '20 at 12:10

Nahuel Fouilleul · Accepted Answer · 2020-09-02T08:22:25.297

1

If the monitoring of the background job is not needed the double fork may be a solution.

 ( ./emulator "@$1" &>/dev/null & )

when the parent exit before the child process the child process is attached to process 1 so no more monitored by current shell.

also looking at the script, cd - could be avoided changing directory in the sub shell:

(
    cd "$ANDROID_HOME/emulator" 
    ./emulator "@$1" &
) &>/dev/null

because current shell will not be affected by subshell change directory

edited Sep 02 '20 at 08:22

answered Sep 02 '20 at 08:11

Nahuel Fouilleul

18,726
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Could you please explain or provide a deeper source for "double fork"? – winklerrr Sep 02 '20 at 08:11
I've found [this question](https://stackoverflow.com/questions/881388/what-is-the-reason-for-performing-a-double-fork-when-creating-a-daemon) you might find useful, that elaborates a bit more about it. – Miguel Sep 02 '20 at 12:09

How to suppress function output in Bash?

2 Answers2