Error when using mpirun with a shell script - shell

When I run
mpirun -np 4 mpi_script.sh
I get the error
Open MPI tried to fork a new process via the "execve" system call but failed.
...
Error: Exec format error
despite the fact that I can run the script with ./mpi_script.sh

In my case the problem was I didn't have a shebang.
Adding #!/usr/bin/env bash to the top of my script fixed it:
#!/usr/bin/env bash
# rest of script
# ...
N.b. be sure that the file has execute permissions:
chmod +x mpi_script.sh

Related

How to debug a tcsh script and put the debugged output in a file?

I want to debug my tcsh script. Like in sh script if we add set command with -x then it prints the execution of script.
#! /bin/sh
set -x
exec 2>/usr/bin/error.log
#My script code
And I tried this as explained here . But still don't know how to print in a file? I am writing below code:
#! /bin/tcsh
set echo
exex 2>/usr/bin/error.log
I am getting permission denied error.I have also tried running as root still same error. But that can be the error in my script. So I want to check in error.log where does the error occur. But after I run script the error.log file is not getting generated. Is there any other way to write stderror for tcsh? Can we use exec command in tcsh?
Also are tcsh and csh both same things or different?

Shell script: unexpected `(' [duplicate]

I have written the following code:
#!/bin/bash
#Simple array
array=(1 2 3 4 5)
echo ${array[*]}
And I am getting error:
array.sh: 3: array.sh: Syntax error: "(" unexpected
From what I came to know from Google, that this might be due to the fact that Ubuntu is now not taking "#!/bin/bash" by default... but then again I added the line but the error is still coming.
Also I have tried by executing bash array.sh but no luck! It prints blank.
My Ubuntu version is: Ubuntu 14.04
Given that script:
#!/bin/bash
#Simple array
array=(1 2 3 4 5)
echo ${array[*]}
and assuming:
It's in a file in your current directory named array.sh;
You've done chmod +x array.sh;
You have a sufficiently new version of bash installed in /bin/bash (you report that you have 4.3.8, which is certainly new enough); and
You execute it correctly
then that should work without any problem.
If you execute the script by typing
./array.sh
the system will pay attention to the #!/bin/bash line and execute the script using /bin/bash.
If you execute it by typing something like:
sh ./array.sh
then it will execute it using /bin/sh. On Ubuntu, /bin/sh is typically a symbolic link to /bin/dash, a Bourne-like shell that doesn't support arrays. That will give you exactly the error message that you report.
The shell used to execute a script is not affected by which shell you're currently using or by which shell is configured as your login shell in /etc/passwd or equivalent (unless you use the source or . command).
In your own answer, you say you fixed the problem by using chsh to change your default login shell to /bin/bash. That by itself should not have any effect. (And /bin/bash is the default login shell on Ubuntu anyway; had you changed it to something else previously?)
What must have happened is that you changed the command you use from sh ./array.sh to ./array.sh without realizing it.
Try running sh ./array.sh and see if you get the same error.
Instead of using sh to run the script,
try the following command:
bash ./array.sh
I solved the problem miraculously. In order to solve the issue, I found a link where it was described to be gone by using the following code. After executing them, the issue got resolved.
chsh -s /bin/bash adhikarisubir
grep ^adhikarisubir /etc/passwd
FYI, "adhikarisubir" is my username.
After executing these commands, bash array.sh produced the desired result.

Script not working as Command line

i've created simple bash script that do the following
:
#!/usr/bin/env bash
cf ssh "$1"
When I run the command line from the CLI like cf ssh myapp its running as expected, but when I run the script like
. myscript.sh myapp
I got error: App not found
I dont understand what is the difference, I've provided the app name after I invoke the script , what could be missing here ?
update
when I run the script with the following its working, any idea why the "$1" is not working ...
#!/usr/bin/env bash
cf ssh myapp
When you do this:
. myscript.sh myapp
You don't run the script, but you source the file named in the first argument. Sourcing means reading the file, so it's as if the lines in the file were typed on the command line. In your case what happens is this:
myscript.sh is treates as the file to source and the myapp argument is ignored.
This line is treated as a comment and skipped.
#!/usr/bin/env bash
This line:
cf ssh "$1"
is read as it stands. "$1" takes the value of $1 in the calling shell. Possibly - most likely in your case - it's blank.
Now you should know why it works as expected when you source this version of your script:
#!/usr/bin/env bash
cf ssh myapp
There's no $1 to resolve, so everything goes smoothly.
To run the script and be able to pass arguments to it, you need to make the file executable and then execute it (as opposed to sourcing). You can execute the script for example this way:
./script.bash arg1 arg2

Jenkins fails with Execute shell script

I have my bash script in ${JENKINS_HOME}/scripts/convertSubt.sh
My job has build step Execute shell:
However after I run job it fails:
The error message (i.e. the 0: part) suggests, that there is an error while executing the script.
You could run the script with
sh -x convertSubt.sh
For the safe side, you could also do a
ls -l convertSubt.sh
file convertSubt.sh
before you run it.
make sure that the script exist with ls
no need to sh , just ./convertSubs.sh ( make sure you have run permissions)

How to source a csh script from inside a bash script

My default shell is bash. I have set some environment variables in my .bashrc file.
I installed a program which use .cshrc file. It contains the path to several cshell scripts.
When I run the following commands in the shell windows it works perfectly :
exec csh
source .cshrc
exec bash
I have tried to put these commands in bash script, unfortunately it didn't work.
is there another way to write a script in order to get the same result as running commands from a shell windows.
I hope my question is now clear
Many thanks for any help
WARNING : don't put the following script in your .bashrc, it will reload bash and so reload .bashrc again and again (stopable with C-c anyway)
Use preferable this script in your kit/CDS stuff startup script. (cadence presumably)
WARNING 2 : if anything in your file2source fails, the whole 'trick' stops.
Call this script : cshWrapper.csh
#! /bin/csh
# to launch using
# exec cshWrapper.csh file2source.sh
source $1
exec $SHELL -i
and launch it using
exec ./cshWrapper.csh file2source.sh
it will : launch csh, source your file and came back to the same parrent bash shell
Example :
$> ps
PID TTY TIME CMD
7065 pts/0 00:00:02 bash
$>exec ./cshWrapper.csh toggle.csh
file sourced
1
$> echo $$
7065
where in my case i use the file toggle.csh
#! /bin/csh
# source ./toggle.csh
if ! $?TOGGLE then
setenv TOGGLE 0
endif
if ($?TOGGLE) then
echo 'file sourced'
if ($TOGGLE == 0) then
setenv TOGGLE 1
else
setenv TOGGLE 0
endif
endif
echo $TOGGLE
Hope it helps
New proposal, since I faced another problem with exec.
exec kills whatever remains in the script, except if you force a fork by using a pipe after it `exec script |cat'. In such case if you have environment variable in the script, they are not spread back to the script itself, which is not what we want. The only solution I found is to use 3 files (let's call them for the example : main.bash that call first.cshrc and second.sh).
#! /bin/bash
#_main.bash_
exec /bin/csh -c "source /path_to_file/cshrc; exec /bin/bash -i -c /path_to_file/second.sh"
# after exec nothing remains (like Attila the Hun)
# the rest of the script is in 'second.sh'
With that manner, i can launch in a single script call, an old cshrc design kit, and still process some bash command after, and finally launch the main program in bash (let say virtuoso)

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