I have a stopped process in Linux at a given terminal. Now I am at another terminal. How do I start that process. What kill signal would I send. I own that process.
You can issue a kill -CONT pid, which will do what you want as long as the other terminal session is still around. If the other session is dead it might not have anywhere to put the output.
In addition to #Dave's answer, there is an advanced method to redirect input and output file descriptors of a running program using GDB.
A FreeBSD example for an arbitrary shell script with PID 4711:
> gdb /bin/sh 4711
...
Attaching to program: /bin/sh, process 4711
...
(gdb) p close(1)
$1 = 0
(gdb) p creat("/tmp/testout.txt",0644)
$2 = 1
(gdb) p close(2)
$3 = 0
(gdb) p dup2(1,2)
$4 = 2
EDIT - explanation: this closes filehandle 1, then opens a file, which reuses 1. Then it closes filehandle 2 and duplicates filehandle 1 to 2.
Now this process' stdout and stderr go to indicated file and are readable from there. If stdin is required, you need to p close(0) and then attach some input file or PIPE or smth.
For the time being, I could not find a method to remotely disown this process from the controlling terminal, which means that when the terminal exits, this process receives SIGHUP signal.
Note: If you do have/gain access to the other terminal, you can disown -a so that this process will continue to run after the terminal closes.
Related
I'm making a custom shell in c, and I'm not entirely sure how processes running in the background works.
I thought I had a pretty good grip, but I'm not sure what happens to file descriptor 0, for example.
A command like 'cat', when passed without arguments expects input from stdin.
If you run cat in the background without arguments (e.g.: cat &), it will immediately terminate. Why does this actually happen, was cat's file descriptor 0 closed somehow?
I tried handling this by setting its file descriptor 0 to /dev/null, but then cat complains about having a bad file descriptor.
Any information about how the shell actually handles stdin to background processes would be much appreciated!
Update: Setting fd 0 to /dev/null works, but is that what the shell actually does? Does the shell set the file descriptor 0 of the background process to /dev/null? If so, what does the shell do if you move to to foreground, does it set it back to 0? (by the way, I'm doing this for the first process in a pipe if there is more than one process)
Here's what happens when I run cat in the background:
$ cat&
[1] 19322
$
[1]+ Stopped cat
$
Note that the cat process has not terminated. It's been suspended.
$ jobs
[1]+ Stopped cat
$ ps u 19322
USER PID %CPU %MEM VSZ RSS TTY STAT START TIME COMMAND
rici 19322 0.0 0.0 11788 784 pts/4 T 01:48 0:00 cat
Backgrounded tasks cannot read from a terminal device. When they try to do that, they receive a SIGTTIN signal. The default action associated with that signal is to stop the task. But it hasn't been terminated; you can still resuscitate it by bringing it back to the foreground:
$ fg
cat
Hello, world!
Hello, world!
There is a detailed explanation in APUE, section 9.8.
I have written a simple application which has two threads. First thread is printing to stdout and second thread is reading from stdin.
I have redirected stdin, stdout and stdin of the process to 3 different pipes as below.
mkfifo pipe_in && mkifo pipe_out && mkfifo pipe_err
./a.out < pipe_in 2> pipe_err 1> pipe_out &
Problem is that I this application (./a.out) is blocked from running until I do below:
cat < pipe_out &
cat < pipe_err &
cat > pipe_in
Why is this application blocked? Is it because no body from other side has opened the pipe?
What is the workaround for the problem where I want to run my application without it being blocked completely. I want only the thread which is waiting for user input to get blocked and other thread to continue execution.
This application is started at bootup. So, this application should be run without getting blocked for user input. User can use anytime "cat > pipe_in" to start providing input to get some details about this application.
Redirection is done by the shell, before starting the application program. Thus a.out does not start, and cannot create any threads that do anything, until the opens of the pipes complete, and the opens of the pipe write sides (for 1 and 2) do not complete until the read sides are opened. (The open of a read side for 0 does complete immediately.)
(OSX 10.7) An application we use let us assign scripts to be called when certain activities occur within the application. I have assigned a bash script and it's being called, the problem is that what I need to do is to execute a few commands, wait 30 seconds, and then execute some more commands. If I have my bash script do a "sleep 30" the entire application freezes for that 30 seconds while waiting for my script to finish.
I tried putting the 30 second wait (and the second set of commands) into a separate script and calling "./secondScript &" but the application still sits there for 30 seconds doing nothing. I assume the application is waiting for the script and all child processes to terminate.
I've tried these variations for calling the second script from within the main script, they all have the same problem:
nohup ./secondScript &
( ( ./secondScript & ) & )
( ./secondScript & )
nohup script -q /dev/null secondScript &
I do not have the ability to change the application and tell it to launch my script and not wait for it to complete.
How can I launch a process (I would prefer the process to be in a scripting language) such that the new process is not a child of the current process?
Thanks,
Chris
p.s. I tried the "disown" command and it didn't help either. My main script looks like this:
[initial commands]
echo Launching second script
./secondScript &
echo Looking for jobs
jobs
echo Sleeping for 1 second
sleep 1
echo Calling disown
disown
echo Looking again for jobs
jobs
echo Main script complete
and what I get for output is this:
Launching second script
Looking for jobs
[1]+ Running ./secondScript &
Sleeping for 1 second
Calling disown
Looking again for jobs
Main script complete
and at this point the calling application sits there for 45 seconds, waiting for secondScript to finish.
p.p.s
If, at the top of the main script, I execute "ps" the only thing it returns is the process ID of the interactive bash session I have open in a separate terminal window.
The value of $SHELL is /bin/bash
If I execute "ps -p $$" it correctly tells me
PID TTY TIME CMD
26884 ?? 0:00.00 mainScript
If I execute "lsof -p $$" it gives me all kinds of results (I didn't paste all the columns here assuming they aren't relevant):
FD TYPE NAME
cwd DIR /private/tmp/blahblahblah
txt REG /bin/bash
txt REG /usr/lib/dyld
txt REG /private/var/db/dyld/dyld_shared_cache_x86_64
0 PIPE
1 PIPE -> 0xffff8041ea2d10
2 PIPE -> 0xffff 8017d21cb
3r DIR /private/tmp/blahblah
4r REG /Volumes/DATA/blahblah
255r REG /Volumes/DATA/blahblah
The typical way of doing this in Unix is to double fork. In bash, you can do this with
( sleep 30 & )
(..) creates a child process, and & creates a grandchild process. When the child process dies, the grandchild process is inherited by init.
If this doesn't work, then your application is not waiting for child processes.
Other things it may be waiting for include the session and open lock files:
To create a new session, Linux has a setsid. On OS X, you might be able to do it through script, which incidentally also creates a new session:
# Linux:
setsid sleep 30
# OS X:
nohup script -q -c 'sleep 30' /dev/null &
To find a list of inherited file descriptors, you can use lsof -p yourpid, which will output something like:
sleep 22479 user 0u CHR 136,32 0t0 35 /dev/pts/32
sleep 22479 user 1u CHR 136,32 0t0 35 /dev/pts/32
sleep 22479 user 2u CHR 136,32 0t0 35 /dev/pts/32
sleep 22479 user 5w REG 252,0 0 1048806 /tmp/lockfile
In this case, in addition to the standard FDs 0, 1 and 2, you also have a fd 5 open with a lock file that the parent can be waiting for.
To close fd 5, you can use exec 5>&-. If you think the lock file might be stdin/stdout/stderr themselves, you can use nohup to redirect them to something else.
Another way is to abandon the child
#!/bin/bash
yourprocess &
disown
As far as I understand, the application replaces the normal bash shell because it is still waiting for a process to finish even if init should have taken care of this child process.
It could be that the "application" intercepts the orphan handling which is normally done by init.
In that case, only a parallel process with some IPC can offer a solution (see my other answer)
I think it depends on how your parent process tries to detect if your child process has been finished.
In my case (my parent process was gnu make), I succeed by closing stdout and stderr (slightly based on the answer of that other guy) like this:
sleep 30 >&- 2>&- &
You might also close stdin
sleep 30 <&- >&- 2>&- &
or additionally disown your child process (not for Mac)
sleep 30 <&- >&- 2>&- & disown
Currently tested only in bash on kubuntu 14.04 and Mac OSX.
If all else fails:
Create a named pipe
start the "slow" script independent from the "application", make sure executes it's task in an endless loop, starting with reading from the pipe. It will become read-blocked when it tries to read..
from the application, start your other script. When it needs to invoke the "slow" script, just write some data to the pipe. The slow script will start independently so your script won't wait for the "slow" script to finish.
So, to answer the question:
bash - how can I launch a new process that is NOT a child of the original process?
Simple: don't launch it but let an independent entity launch it during boot...like init or on the fly with the command at or batch
Here I have a shell
└─bash(13882)
Where I start a process like this:
$ (urxvt -e ssh somehost&)
I get a process tree (this output snipped from pstree -p):
├─urxvt(14181)───ssh(14182)
where the process is parented beneath pid 1 (systemd in my case).
However, had I instead done this (note where the & is) :
$ (urxvt -e ssh somehost)&
then the process would be a child of the shell:
└─bash(13882)───urxvt(14181)───ssh(14182)
In both cases the shell prompt is immediately returned and I can exit
without terminating the process tree that I started above.
For the latter case the process tree is reparented beneath pid 1 when
the shell exits, so it ends up the same as the first example.
├─urxvt(14181)───ssh(14182)
Either way, the result is a process tree that outlives the shell. The
only difference is the initial parenting of that process tree.
For reference, you can also use
nohup urxvt -e ssh somehost &
urxvt -e ssh somehost & disown $!
Both give the same process tree as the second example above.
└─bash(13882)───urxvt(14181)───ssh(14182)
When the shell is terminated the process tree is, like before, reparented
to pid 1.
nohup additionally redirects the process' standard output to a file
nohup.out so, if that is a useful trait, it may be a more useful choice.
Otherwise, with the first form above, you immediately have a completely
detached process tree.
I current have a script that looks like this.
# code
mplayer "$vid"
# more code
The problem is that if this script is killed the mplayer process lives. I wondering how I could make it so that killing the script would kill mplayer as well.
I can't use exec because I need to run commands after mplayer.
exec mplayer "$vid"
The only possible solution I can think of is to spawn it in the background and wait until it finishes manually. That way I can get it's PID and kill it when the script gets killed, not exactly elegant. I was wondering what the "proper" or best way of doing this is.
I was able to test the prctl idea I posted about in a comment and it seems to work. You will need to compile this:
#include "sys/prctl.h"
#include "stdlib.h"
#include "string.h"
#include "unistd.h"
int main(int argc, char ** argv){
prctl(PR_SET_PDEATHSIG, atoi(argv[1]),0,0,0);
char * argv0 = strdup(argv[2]);
char * slashptr = strrchr(argv0, '/');
if(slashptr){
argv0 = slashptr + 1;
}
return execvp(argv0, &(argv[2]));
}
Let's say you have compiled the above to an executable named "prun" and it is in your path. Let's say your script is called "foo.sh" and it is also in your path. Make a wrapper script that calls
prun 15 foo.sh
foo.sh should get SIGTERM when the wrapper script is terminated for any reason, even SIGKILL.
Note: this is a linux only solution and the c source code presented is without detailed checking of arguments
Thanks to Mux for the lead. It appears that there is no way to do this in bash except for manually catching signals. Here is a final working (overly commented) version.
trap : SIGTERM SIGINT # Trap these two (killing) signals. These will cause wait
# to return a value greater than 128 immediately after received.
mplayer "$vid" & # Start in background (PID gets put in `$!`)
pid=$!
wait $pid # Wait for mplayer to finish.
[ $? -gt 128 ] && { kill $pid ; exit 128; } ; # If a signal was recieved
# kill mplayer and exit.
Refrences:
- traps: http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_12_02.html
(Updated) I think I understand what you are looking for now:
You can accomplish this by spawning a new terminal to run your script:
gnome-terminal -x /path_to_dir_of_your_script/your_script_name
(or use xterm -e or konsole -e instead of gnome-terminal -x, depending on what system you are on)
So now whenever your script ends / exits (I assume you have exit 0 or exit 1 in certain parts of the script), the newly spawned terminal will also exit since the script is finished - this will in turn also kill any applications spawned under that new terminal.
For example, I just tested the above command with this script:
#!/bin/bash
gedit &
pid=$!
echo "$pid"
sleep 5
exit 0
As you can see, there are no explicit calls to kill the new gedit process, but the application (gedit) closes as soon as the script exits anyway.
(Previous answer: alternatively, if you were simply asking about how to kill a process) Here's a short example of how you can accomplish that with kill.
#!/bin/bash
gedit &
pid=$!
echo "$pid"
sleep 5
kill -s SIGKILL $pid
Unless I misunderstood your question, you can get the PID of the spawned process right away instead of waiting until it finishes.
Well, you can simply kill the process group instead, this way the whole process tree will be killed, first find out the group id
ps x -o "%p %r %c" | grep <name>
And then use kill like so:
kill -TERM -<gid>
Note the dash before the process group id. Or a one-liner:
kill -TERM -$(pgrep <name>)
Perhaps use command substitution to run mplayer "$vid" in a subshell:
$(mplayer "$vid")
I tested it this way:
tesh.sh:
#!/bin/sh
$vid = "..."
$(mplayer "$vid")
% test.sh
In a separate terminal:
% pkill test.sh
In the orginal terminal, mplayer stops, printing to stderr
Terminated
MPlayer interrupted by signal 13 in module: av_sync
I've executed the following C code in Linux CentOS to create a process.
#include <stdio.h>
#include <unistd.h>
int main ()
{
int i = 0;
while ( 1 )
{
printf ( "\nhello %d\n", i ++ );
sleep ( 2 );
}
}
I've compiled it to hello_count. When I do ./hello_count in a terminal window, The output is like this:
hello 0
hello 1
hello 2
...
When I enter the command in another terminal window
ps -e
the process 2956 ./hello_count is listed there. In the same window, I've stopped the process using:
kill -s SIGSTOP 2956
When I enter the following command again,
ps -e
the process 2956 ./hello_count is still listed.
Then I've entered the following command to resume the process in the same window itself.
kill -s SIGCONT 2956
However, the process resumed in the previous window in which it was executing and giving output.
Is there any command or any method to resume (not to restart) the process with pid 2956 in a different terminal window?
I mean, I need the output like,
hello 8
hello 9
...
in a window other than the one in which I was getting the above output before I've stopped the process.
The problem you're having is that your process is attached to a particular tty and switching tty once a process is started isn't normally possible. See this question.
There are some hacky methods you could consider mind you.
For real world command line scenarios, using screen would allow you to start a command in a virtual terminal and then connect to that terminal from any other. But this isn't a programatic solution which your question seems to indicate you're looking for.