For example, when I type ping &, why can we see ping process working? And when I type find / -name '*test*' & , it's also seen on monitor.

Why is that? Is it not really in background?


The & directs the shell to run the command in the background, i.e, it is forked and run in a separate sub-shell, as a job, asynchronously.

Note that when you put & the output - both stdout and stderr - will still be printed onto the screen. If you do not want to see any output on the screen, redirect both stdout and stderr to a file by:

myscript > ~/myscript.log 2>&1 &

Usually you may want to discard the stderr by redirecting it to /dev/null if you are not worried about analyzing errors later.

You can also even run commands/scripts at the same time, in separate sub-shells. For eg;

./script1 & ./script2 & ./script3 & 

A background job can be brought back to the command line before it finishes with the command:

fg <job-number>

The job-number can be obtained by running

  • Good answer. You could include that one can also use /dev/null as a file to redirect terminal output if one wants to discard the output instead of saving it somewhere. And am I right that 2>&1 redirects stderr to stdout? Maybe you want to clarify that too... Thanks. – Byte Commander Aug 18 '15 at 9:44
  • @ByteCommander Thanks :) edited, hope it is clear now. Yes, you are right, 2>&1 redirects stderr to stdout. – Ron Aug 18 '15 at 10:27
  • 3
    I think it would be worth mentioning that stty tostop will cause background jobs to be suspended if they try to write to the terminal. – kasperd Aug 18 '15 at 11:12
  • cmd_with_params >/dev/null 2>/dev/null & – tgkprog May 22 '17 at 16:11

When you use &, the process is running in background. But its standard output is still the terminal.

In fact, you may run ping & and find / -name '*test*' & at the same time (resulting in a mixed output), but you may not run ping and find / -name '*test*' at the same time on the same shell.

If you don't want to see anything, use something like ping &> /dev/null &.

Additionally, you may want to learn about nohup and disown.

  • disown doesn't prevent output from being printed to the terminal. – Ruslan Aug 18 '15 at 11:53
  • @Ruslan: no, but it removes the job from the shell. Which may (or may not) be something the OP may be interested in. – Andrea Corbellini Aug 18 '15 at 11:55

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