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How do you catch error codes in a shell pipe?
(5 answers)
Closed 7 years ago.
I have a pipeline, say a|b where if a runs into a problem, I want to stop the whole pipeline.
'a exiting with exit=1 doesn't do this as often 'b doesn't care about return codes.
e.g.
echo 1|grep 0|echo $? <-- this shows that grep did exit=1
but
echo 1|grep 0 | wc <--- wc is unfazed by grep's exit here
If I ran the pipeline as a subprocess of an owning process, any of the pipeline processes could kill the owning process. But this seems a bit clumsy -- but it would zap the whole pipeline.
Not possible with basic shell constructs, probably not possible in shell at all.
Your first example doesn't do what you think. echo doesn't use standard input, so putting it on the right side of a pipe is never a good idea. The $? that you're echoing is not the exit value of the grep 0. All commands in a pipeline run simultaneously. echo has already been started, with the existing value of $?, before the other commands in the pipeline have finished. It echoes the exit value of whatever you did before the pipeline.
# The first command is to set things up so that $? is 2 when the
# second command is parsed.
$ sh -c 'exit 2'
$ echo 1|grep 0|echo $?
2
Your second example is a little more interesting. It's correct to say that wc is unfazed by grep's exit status. All commands in the pipeline are children of the shell, so their exit statuses are reported to the shell. The wc process doesn't know anything about the grep process. The only communication between them is the data stream written to the pipe by grep and read from the pipe by wc.
There are ways to find all the exit statuses after the fact (the linked question in the comment by shx2 has examples) but a basic rule that you can't avoid is that the shell will always wait for all the commands to finish.
Early exits in a pipeline sometimes do have a cascade effect. If a command on the right side of a pipe exits without reading all the data from the pipe, the command on the left of that pipe will get a SIGPIPE signal the next time it tries to write, which by default terminates the process. (The 2 phrases to pay close attention to there are "the next time it tries to write" and "by default". If a the writing process spends a long time doing other things between writes to the pipe, it won't die immediately. If it handles the SIGPIPE, it won't die at all.)
In the other direction, when a command on the left side of a pipe exits, the command on the right side of that pipe gets EOF, which does cause the exit to happen fairly soon when it's a simple command like wc that doesn't do much processing after reading its input.
With direct use of pipe(), fork(), and wait3(), it would be possible to construct a pipeline, notice when one child exits badly, and kill the rest of them immediately. This requires a language more sophisticated than the shell.
I tried to come up with a way to do it in shell with a series of named pipes, but I don't see it. You can run all the processes as separate jobs and get their PIDs with $!, but the wait builtin isn't flexible enough to say "wait for any child in this set to exit, and tell me which one it was and what the exit status was".
If you're willing to mess with ps and/or /proc you can find out which processes have exited (they'll be zombies), but you can't distinguish successful exit from any other kind.
Write
set -e
set -o pipefail
at the beginning of your file.
-e will exit on an error and -o pipefail will produce an errorcode on each stage of you "pipeline"
Related
First of all, sorry if the title is not clear or misleading, my question is not not exactly easy to be understood out of context.
So here it is: I am running a shell script (hello.sh) that needs to relocate itself from /root to /.
thus I made a simple recursion, to test from where the script is running and to make a temporary copy and launch it and exit(this last temporary copy will move the original file, and delete itself while still running).
#!/bin/sh
IsTMP=$(echo $0 | grep "tmp")
if [ -z "$IsTMP" ]; then
cp /root/hello.sh /tmp/hello.sh
/bin/sh /tmp/hello.sh &
exit
else
unlink /hello.sh
rsync /root/hello.sh /hello.sh
rm /root/hello.sh
rm /tmp/hello.sh
fi
while true; do
sleep 5
echo "Still Alive"
done
This script works totally well and suits my needs (even though it is a horrendous hack): the script is moved, and re-executed from a temporary place. However, when i pipe the shell script with a tee, just like:
/hello.sh | tee -a /log&
The behaviour is not the same:
hello.sh is exiting but not tee
When i try to kill tee, the temporary copy is automatically killed after a few seconds, without entering the infinite loop
This behaviour is the exact same if i replace tee with another binary (e.g. watch,...), so I am wondering if it comes from piping.
Sorry if i am not too clear about my problem.
Thanks in advance.
When i try to kill tee, the temporary copy is automatically killed after a few seconds, without entering the infinite loop
That's not the case. The script is entering the infinite loop, the few seconds are the five the sleep 5 in the loop pauses, and then it is killed by the signal SIGPIPE (Broken pipe) because it tries to echo "Still Alive" to the pipe which is closed on the read end since tee has been killed.
There is no link between tee and the second instance
That's not the case. There is a link, namely the pipe, the write end of which is the standard output of the parent as well as (inherited) the child shell script, and the read end is the standard input of tee. You can see this if you look at ls -l /proc/pid/fd, where pid is the process id of the script's shell on the one hand, and of tee on the other.
Good morning all,
I am trying to implement concurrency in a very specific environment, and keep getting stuck. Maybe you can help me.
this is the situation:
-I have N nodes that can read/write in a shared folder.
-I want to execute an application in one of them. this can be anything, like a shell script, an installed code, or whatever.
-To do so, I have to send the same command to all of them. The first one should start the execution, and the rest should see that somebody else is running the desired application and exit.
-The execution of the application can be killed at any time. This is important because does not allow relying on any cleaning step after the execution.
-if the application gets killed, the user may want to execute it again. He would then send the very same command.
My current approach is to create a shell script that wraps the command to be executed. This could also be implemented in C. Not python or other languages, to avoid library dependencies.
#!/bin/sh
# (folder structure simplified for legibility)
mutex(){
lockdir=".lock"
firstTask=1 #false
if mkdir "$lockdir" &> /dev/null
then
controlFile="controlFile"
#if this is the first node, start coordinator
if [ ! -f $controlFile ]; then
firstTask=0 #true
#tell the rest of nodes that I am in control
echo "some info" > $controlFile
fi
# remove control File when script finishes
trap 'rm $controlFile' EXIT
fi
return $firstTask
}
#The basic idea is that a task executes the desire command, stated as arguments to this script. The rest do nothing
if ! mutex ;
then
exit 0
fi
#I am the first node and the only one reaching this, so I execute whatever
$#
If there are no failures, this wrapper works great. The problem is that, if the script is killed before the execution, the trap is not executed and the control file is not removed. Then, when we execute the wrapper again to restart the task, it won't work as every node will think that somebody else is running the application.
A possible solution would be to remove the control script just before the "$#" call, but that it would lead to some race condition.
Any suggestion or idea?
Thanks for your help.
edit: edited with correct solution as future reference
Your trap syntax looks wrong: According to POSIX, it should be:
trap [action condition ...]
e.g.:
trap 'rm $controlFile' HUP INT TERM
trap 'rm $controlFile' 1 2 15
Note that $controlFile will not be expanded until the trap is executed if you use single quotes.
I am running a shell script, something like sh script.sh in bash. The script contains many lines, some of which take seconds and others take days to execute. How can I kill the sh command but not kill its command currently running (the current line from the script)?
You haven't specified exactly what should happen when you 'kill' your script., but I'm assuming that you'd like the currently executing line to complete and then exit before doing any more work.
This is probably best achieved only by coding your script to behave in such a way as to receive such a kill command and respond in an appropriate way - I don't think that there is any magic to do this in linux.
for example:
You could trap a signal and then set a variable
Check for existence of a file (e.g touch /var/tmp/trigger)
Then after each line in your script, you'd need to check to see if each the trap had been called (or your trigger file created) - and then exit. If the trigger has not been set, then you continue on and do the next piece of work.
To the best of my knowledge, you can't trap a SIGKILL (-9) - if someone sends that to your process, then it will die.
HTH, Ace
The only way I can think of achieving this is for the parent process to trap the kill signal, set a flag, and then repeatedly check for this flag before executing another command in your script.
However the subprocesses need to also be immune to the kill signal. However bash seems to behave different to ksh in this manner and the below seems to work fine.
#!/bin/bash
QUIT=0
trap "QUIT=1;echo 'term'" TERM
function terminated {
if ((QUIT==1))
then
echo "Terminated"
exit
fi
}
function subprocess {
typeset -i N
while ((N++<3))
do
echo $N
sleep 1
done
}
while true
do
subprocess
terminated
sleep 3
done
I assume you have your script running for days and then you don't just want to kill it without knowing if one of its children finished.
Find the pid of your process, using ps.
Then
child=$(pgrep -P $pid)
while kill -s 0 $child
do
sleep 1
done
kill $pid
I followed this blog entry to parallelize sort by splitting a large file, sorting and merging.
The steps are:
split -l5000000 data.tsv '_tmp'
ls -1 _tmp* | while read FILE; do sort $FILE -o $FILE & done
sort -m _tmp* -o data.tsv.sorted
Between step 2 and 3, one must wait until the sorting step has finished.
I assumed that wait without any arguments would be the right thing, since according to the man page, if wait is called without arguments all currently active child processes are waited for.
However, when I try this in the shell (i.e. executing steps 1 and 2, and then wait), wait returns immediately, although top shows the sort processes are still running.
Ultimately I want to increase the speed of a script with that, so its not a one time thing I could do manually on the shell.
I know sort has a --parallel option since version 8, however on the cluster I am running this, an older version is installed, and I am also curious about how to solve this issue.
Here's a simple test case reproducing your problem:
true | { sleep 10 & }
wait
echo "This echos immediately"
The problem is that the pipe creates a subshell, and the forked processes are part of that subshell. The solution is to wait in that subshell instead of your main parent shell:
true | { sleep 10 & wait }
echo "This waits"
Translated back into your code, this means:
ls -1 _tmp* | { while read FILE; do sort $FILE -o $FILE & done; wait; }
From the bash man page:
Each command in a pipeline is executed as a separate process (i.e., in a subshell).
So when you pipe to while, a subshell is created. Everything else in step 2 is executed within this subshell, (ie, all the background processes). The script then exits the while loop, leaving the subshell, and wait is executed in the parent shell, where there is nothing to wait for. You can avoid using the pipeline by using a process substitution:
while read FILE; do
sort $FILE -o $FILE &
done < <(ls -1 _tmp*)
For some reasons not relevant to this question, I am running a Java server in a bash script not directly but via command substitution under a separate sub-shell, and in the background. The intent is for the subcommand to return the process id of the Java server as its standard output. The fragement in question is as follows:
launch_daemon()
{
/bin/bash <<EOF
$JAVA_HOME/bin/java $JAVA_OPTS -jar $JAR_FILE daemon $PWD/config/cl.yml <&- &
pid=\$!
echo \${pid} > $PID_FILE
echo \${pid}
EOF
}
daemon_pid=$(launch_daemon)
echo ${daemon_pid} > check.out
The Java daemon in question prints to standard error and quits if there is a problem in initialization, otherwise it closes standard out and standard err and continues on its way. Later in the script (not shown) I do a check to make sure the server process is running. Now on to the problem.
Whenever I check the $PID_FILE above, it contains the correct process id on one line.
But when I check the file check.out, it sometimes contains the correct id, other times it contains the process id repeated twice on the same line separated by a space charcater as in:
34056 34056
I am using the variable $daemon_pid in the script above later on in the script to check if the server is running, so if it contains the pid repeated twice this totally throws off the test and it incorrectly thinks the server is not running. Fiddling with the script on my server box running CentOS Linux by putting in more echo statements etc. seems to flip the behavior back to the correct one of $daemon_pid containing the process id just once, but if I think that has fixed it and check in this script to my source code repo and do a build and deploy again, I start seeing the same bad behavior.
For now I have fixed this by assuming that $daemon_pid could be bad and passing it through awk as follows:
mypid=$(echo ${daemon_pid} | awk '{ gsub(" +.*",""); print $0 }')
Then $mypid always contains the correct process id and things are fine, but needless to say I'd like to understand why it behaves the way it does. And before you ask, I have looked and looked but the Java server in question does NOT print its process id to its standard out before closing standard out.
Would really appreciate expert input.
Following the hint by #WilliamPursell, I tracked this down in the bash source code. I honestly don't know whether it is a bug or not; all I can say is that it seems like an unfortunate interaction with a questionable use case.
TL;DR: You can fix the problem by removing <&- from the script.
Closing stdin is at best questionable, not just for the reason mentioned by #JonathanLeffler ("Programs are entitled to have a standard input that's open.") but more importantly because stdin is being used by the bash process itself and closing it in the background causes a race condition.
In order to see what's going on, consider the following rather odd script, which might be called Duff's Bash Device, except that I'm not sure that even Duff would approve: (also, as presented, it's not that useful. But someone somewhere has used it in some hack. Or, if not, they will now that they see it.)
/bin/bash <<EOF
if (($1<8)); then head -n-$1 > /dev/null; fi
echo eight
echo seven
echo six
echo five
echo four
echo three
echo two
echo one
EOF
For this to work, bash and head both have to be prepared to share stdin, including sharing the file position. That means that bash needs to make sure that it flushes its read buffer (or not buffer), and head needs to make sure that it seeks back to the end of the part of the input which it uses.
(The hack only works because bash handles here-documents by copying them into a temporary file. If it used a pipe, it wouldn't be possible for head to seek backwards.)
Now, what would have happened if head had run in the background? The answer is, "just about anything is possible", because bash and head are racing to read from the same file descriptor. Running head in the background would be a really bad idea, even worse than the original hack which is at least predictable.
Now, let's go back to the actual program at hand, simplified to its essentials:
/bin/bash <<EOF
cmd <&- &
echo \$!
EOF
Line 2 of this program (cmd <&- &) forks off a separate process (to run in the background). In that process, it closes stdin and then invokes cmd.
Meanwhile, the foreground process continues reading commands from stdin (its stdin fd hasn't been closed, so that's fine), which causes it to execute the echo command.
Now here's the rub: bash knows that it needs to share stdin, so it can't just close stdin. It needs to make sure that stdin's file position is pointing to the right place, even though it may have actually read ahead a buffer's worth of input. So just before it closes stdin, it seeks backwards to the end of the current command line. [1]
If that seek happens before the foreground bash executes echo, then there is no problem. And if it happens after the foreground bash is done with the here-document, also no problem. But what if it happens while the echo is working? In that case, after the echo is done, bash will reread the echo command because stdin has been rewound, and the echo will be executed again.
And that's precisely what is happening in the OP. Sometimes, the background seek completes at just the wrong time, and causes echo \${pid} to be executed twice. In fact, it also causes echo \${pid} > $PID_FILE to execute twice, but that line is idempotent; had it been echo \${pid} >> $PID_FILE, the double execution would have been visible.
So the solution is simple: remove <&- from the server start-up line, and optionally replace it with </dev/null if you want to make sure the server can't read from stdin.
Notes:
Note 1: For those more familiar with bash source code and its expected behaviour than I am, I believe that the seek and close takes place at the end of case r_close_this: in function do_redirection_internal in redir.c, at approximately line 1093:
check_bash_input (redirector);
close_buffered_fd (redirector);
The first call does the lseek and the second one does the close. I saw the behaviour using strace -f and then searched the code for a plausible looking lseek, but I didn't go to the trouble of verifying in a debugger.