bash: silently kill background function process

Question

shell gurus,

I have a bash shell script, in which I launch a background function, say foo(), to display a progress bar for a boring and long command:

foo() {     while [ 1 ]     do         #massively cool progress bar display code         sleep 1     done }  foo & foo_pid=$!  boring_and_long_command kill $foo_pid >/dev/null 2>&1 sleep 10 

now, when foo dies, I see the following text:

/home/user/script: line XXX: 30290 Killed                  foo 

This totally destroys the awesomeness of my, otherwise massively cool, progress bar display.

How do I get rid of this message?

Answer 1

kill $foo_pid wait $foo_pid 2>/dev/null 

BTW, I don't know about your massively cool progress bar, but have you seen Pipe Viewer (pv)? http://www.ivarch.com/programs/pv.shtml

Answer 2

Just came across this myself, and realised "disown" is what we are looking for.

foo & foo_pid=$! disown  boring_and_long_command kill $foo_pid sleep 10 

The death message is being printed because the process is still in the shells list of watched "jobs". The disown command will remove the most recently spawned process from this list so that no debug message will be generated when it is killed, even with SIGKILL (-9).