So I have a job running in Background on my unix terminal, I will have to close the ssh terminal after some time and my job has to continue after that as well. How can I use the nohup or any other possible option to achieve this.
nohup starts a new process. You cannot retroactively apply it to a process that you've already started.
However, if the shell from which you launched the job is bash, ksh, or zsh then the disown job-control builtin may provide what you want. It can either remove the job from job control altogether or just flag the job to not be sent a SIGHUP when the parent shell itself receives one. This is similar, but not necessarily identical, to the effect of starting a process via the nohup command.
Note well that your job may still have issues if any of its standard streams is connected to the session's terminal. That's something that nohup typically clobbers preemptively, but disown cannot modify after the fact. You're normally better off anticipating this need and starting the process with nohup, but if you're not so foresightful then disown is probably your next best bet.
Note also that as a job-control command, disown takes a jobspec to identify the job to operate on, not a process ID. If necessary, you can use the jobs builtin to help determine the appropriate jobspec.
I'm a beginner in Linux, just have some questions about job and process group.
My textbook says 'Unix shells use the abstraction of a job to represent the processes that are created as a result of evaluating a single command line. At any point in time, there is at most one foreground job and zero or more background jobs.
Lets say we have this simple shell code(I leave out some unimportant code, i.e setup argv etc):
When we type the first commnad, for example:./exampleProgram &
Q1- Is a job created? if yes, what process does the job contain?
the main() of shellex.c is invoked, so when execute the line 15: fork() create a new child process, lets say the parent process is p1, and the newly created child process is c1
and if it is background job, then we can type another command in the prompt, ./anotherProgram & and type enter
Q2- I'm pretty sure it just p1 and c1 at the moment and p1 is executing the second command, when it execute the line 15: fork() again,
we have p1, c1 and new created c2, is my understanding correct?
Q3- How many job is there? is it just one job that contains p1, c1, and c2?
Q4- if it is only one job, so when we keeps typing new commands we will only have one job contains one parent process p1 and many child processes c1, c2, c3, c4...
so why my textbook says that Shell can have more than one job? and why there is at most one foreground job and zero or more background jobs.
There is quite a bit to say on this topic, some of which can fit in an answer, and most of which will require further reading.
For Q1, I would say conceptually yes, but jobs are not automatic, and job tracking and control are not magical. I don't see any logic in the code snippits you've show that e.g. establishes and maintains a jobs table. I understand it's just a sample, so maybe the job control logic is elsewhere. Job control is a feature of common, existing Unix shells, but if a person writes a new Unix shell from scratch, job control features would need to be added, as code / logic.
For Q2, the way you've put it is not how I would put it. After the first call to fork(), yes there is a p1 and a c1, but recognize that at first, p1 and c1 are different instances of the same program (shellex); only after the call to execve() is exampleProgram running. fork() creates a child instance of shellex, and execve() causes the child instance of shellex to be replaced (in RAM) by exampleProgram (assuming that's the value of argv[0]).
There is no real sense in which the parent is "executing" the child, nor the process that replaces the child upon execve(), except just to get them going. The parent starts the child and might wait for the child execution to complete, but really a parent and its whole hierarchy of child processes are all executing each on its own, being executed by the kernel.
But yes, if told that the program to run should be run in the background, then shellex will accept further input, and upon the next call to fork(), there will be the parent shellex with two child processes. And again, at first the child c2 will be an instance of shellex, quickly replaced via execve() by whatever program has been named.
(Regarding running in the background, whether or not & has that effect depends upon the logic inside the function named parseline() in the sample code. Shells I'm familiar with use & to say "run this in the background", but there is nothing special nor magical about that. A newly-written Unix shell can do it some other way, with a trailing +, or a leading BG:, or whatever the shell author decides to do.
For Q3 and Q4, the first thing to recognize is that the parent you are calling p1 is the shell program that you've shown. So, no, p1 would not be part of the job.
In Unix, a job is a collection of processes that execute as part of a single pipeline. Thus a job can consist of one process or many. Such processes remain attached to the terminal from which they are run, but might be in the foreground (running and interactive), suspended, or in the background (running, not interactive).
one process, foreground : ls -lR
one process, background : ls -lR &
one process, background : ls -lR, then CTRL-Z, then bg
many processes, foreground : ls -lR | grep perl | sed 's/^.*\.//'
many processes, background : ls -lR | grep perl | sed 's/^.*\.//' &
To see jobs vs. processes empirically, run a pipeline in the background (the 5th of the 5 examples above), and while it is running use ps to show you the process IDs and the process group IDs. e.g., on my Mac's version of bash, that's:
$ ls -lR | grep perl | sed 's/^.*\.//' &
[1] 2454 <-- job 1, PID of the sed is 2454
$ ps -o command,pid,pgid
COMMAND PID PGID
vim 2450 2450 <-- running in a different tab
ls -lR 2452 2452 }
grep perl 2453 2452 }-- 3 PIDs, 1 PGID
sed s/^.*\.// 2454 2452 }
In contrast to this attachment to the shell and the terminal, a daemon detaches from both. When starting a daemon, the parent uses fork() to start a child process, but then exits, leaving only the child running, and now with a parent of PID 1. The child closes down stdin, stdout, and stderr, since those are meaningless, since a daemon runs "headless".
But in a shell, the parent -- which, again, is the shell -- stays running either wait()ing (foreground child program), or not wait()ing (background child program), and the child typically retains use of stdin, stdout, and stderr (although, these might be redirected to files, etc.)
And, a shell can invoke sub-shells, and of course any program that is run can fork() its own child processes, and so on. So the hierarchy of processes can become quite deep. Without specific action otherwise, a child process will be in the same process group as it's parent.
Here are some articles for further reading:
What is difference between a job and a process in Unix?
https://unix.stackexchange.com/questions/4214/what-is-the-difference-between-a-job-and-a-process
https://unix.stackexchange.com/questions/363126/why-is-process-not-part-of-expected-process-group
Bash Reference Manual; Job Control
Bash Reference Manual; Job Control Basics
A job is not a Linux thing, it's not a background process, it's something your particular shell defines to be a "job".
Typically a shell introduces the notion of "job" to do job control. This normally includes a way to identify a job and perform actions on it, like
bring into foreground
put into background
stop
resume
kill
If a shell has no means to do any of this, it makes little sense to talk about jobs.
I have two programs in Linux (shell scripts, for example):
NeverEnding.sh
AllwaysEnds.sh
The first one does never stop, so I wanna run it in background.
The second one does stop with no problem.
I would like to make a Linux shell script that calls them both, but automatically stops (kill, for example) the first when the second will have finished.
Specific command-line tools allowed, if needed.
You can send the first into the background with & and get the PID of it by $!. Then after the second finishes in the foreground you can kill the first:
#!/bin/bash
NeverEnding.sh &
pid=$!
AllwaysEnds.sh
kill $pid
You don't actually need to save the pid in a variable, since $! only gets updated when you start a background process, it's just make it more easy to read.
I was wondering if there is some way to force to use some specific process ID to Linux to some application before running it. I need to know in advance the process ID.
Actually, there is a way to do this. Since kernel 3.3 with CONFIG_CHECKPOINT_RESTORE set(which is set in most distros), there is /proc/sys/kernel/ns_last_pid which contains last pid generated by kernel. So, if you want to set PID for forked program, you need to perform these actions:
Open /proc/sys/kernel/ns_last_pid and get fd
flock it with LOCK_EX
write PID-1
fork
Voilà! Child will have PID that you wanted.
Also, don't forget to unlock (flock with LOCK_UN) and close ns_last_pid.
You can checkout C code at my blog here.
As many already suggested you cannot set directly a PID but usually shells have facilities to know which is the last forked process ID.
For example in bash you can lunch an executable in background (appending &) and find its PID in the variable $!.
Example:
$ lsof >/dev/null &
[1] 15458
$ echo $!
15458
On CentOS7.2 you can simply do the following:
Let's say you want to execute the sleep command with a PID of 1894.
sudo echo 1893 > /proc/sys/kernel/ns_last_pid; sleep 1000
(However, keep in mind that if by chance another process executes in the extremely brief amount of time between the echo and sleep command you could end up with a PID of 1895+. I've tested it hundreds of times and it has never happened to me. If you want to guarantee the PID you will need to lock the file after you write to it, execute sleep, then unlock the file as suggested in Ruslan's answer above.)
There's no way to force to use specific PID for process. As Wikipedia says:
Process IDs are usually allocated on a sequential basis, beginning at
0 and rising to a maximum value which varies from system to system.
Once this limit is reached, allocation restarts at 300 and again
increases. In Mac OS X and HP-UX, allocation restarts at 100. However,
for this and subsequent passes any PIDs still assigned to processes
are skipped
You could just repeatedly call fork() to create new child processes until you get a child with the desired PID. Remember to call wait() often, or you will hit the per-user process limit quickly.
This method assumes that the OS assigns new PIDs sequentially, which appears to be the case eg. on Linux 3.3.
The advantage over the ns_last_pid method is that it doesn't require root permissions.
Every process on a linux system is generated by fork() so there should be no way to force a specific PID.
From Linux 5.5 you can pass an array of PIDs to the clone3 system call to be assigned to the new process, up to one for each nested PID namespace, from the inside out. This requires CAP_SYS_ADMIN or (since Linux 5.9) CAP_CHECKPOINT_RESTORE over the PID namespace.
If you a not concerned with PID namespaces use an array of size one.
I'm rushing against the clock for a programming assignment in which I have to run a number of instances of the same program on the same machine at the same time.
Currently, I'm starting the instances one at a time, pressing Ctrl+z to pause them, and then doing 'bg %#' to resume execution in the background.
This is extremely tedious and time consuming to do every time I need to test a small change in my application, so I want to write a bash script that will start the multiple instances for me, however I don't know how to do the background switching in a script.
Can anybody please tell me how I can write a simple script that will start a long standing command, pause it, and resume it in the background?
Thanks
Do you want to just start it in the background? For example:
mycommand &
If you want finer grained job control, you can emulate Ctrl-Z and bg. Control-Z sends SIGTSTP ("tty stop") to the program, which suspends it:
kill -TSTP [processid]
And the bg command just sends it a SIGCONT:
kill -CONT [processid]
You don't. You put an ampersand after the command.
command1 &
command2 &
command3 &