bash - Linux process in background - "Stopped" in jobs?

Question

I'm currently running a process with the & sign.

$ example &

However, (please note i'm a newbie to Linux) I realised that pretty much a second after such command I'm getting a note that my process received a stopped signal. If I do

$ jobs

I'll get the list with my example process with a little note "Stopped". Is it really stopped and not working at all in the background? How does it exactly work? I'm getting mixed info from the Internet.

Answer 1

In Linux and other Unix systems, a job that is running in the background, but still has its stdin (or std::cin) associated with its controlling terminal (a.k.a. the window it was run in) will be sent a SIGTTIN signal, which by default causes the program to be completely stopped, pending the user bringing it to the foreground (fg %job or similar) to allow input to actually be given to the program. To avoid the program being paused in this way, you can either:

1. Make sure the programs stdin channel is no longer associated with the terminal, by either redirecting it to a file with appropriate contents for the program to input, or to /dev/null if it really doesn't need input - e.g. myprogram < /dev/null &.
2. Exit the terminal after starting the program, which will cause the association with the program's stdin to go away. But this will cause a SIGHUP to be delivered to the program (meaning the input/output channel experienced a "hangup") - this normally causes a program to be terminated, but this can be avoided by using nohup - e.g. nohup myprogram &.

If you are at all interested in capturing the output of the program, this is probably the best option, as it prevents both of the above signals (as well as a couple others), and saves the output for you to look at to determine if there are any issues with the programs execution:

nohup myprogram < /dev/null > ${HOME}/myprogram.log 2>&1 &