command invoking new shell and script is broken - bash

I have two commands:
/aka/ball/barry/pet elephant/rhino
which invoke new shell and I have command
cleartool setview "/view/epp/lpp/tpp.sh" $VIEW
which also invoke new shell. Is it possible to run both in a single script one by one?

Probably; although it depends. If these shells are designed to be interactive, you can still probably make them work by sending them commands to end the subshells, such as exit 0:
$ /aka/ball/barry/pet elephant/rhino <<< 'exit 0'
If that successfully exits the shell, you can just do it in serial:
#!/bin/bash
##
# Your script?
…stuff
/aka/ball/barry/pet elephant/rhino <<< 'exit 0'
cleartool setview "/view/epp/lpp/tpp.sh" "$VIEW" <<< 'exit 0'
…morestuff
Without knowing more about the nature of the shells these programs invoke, it's hard to say for sure.

Related

What's the effect of combining exec and & in shell script?

I maintained some legacy Linux shell script codes, and I met something like this:
#!/bin/sh
foo()
{
exec some_shell_command &
return 0
}
foo
I'm very curious about the effect of such shell scripts. Is some_shell_command executed in another subprocess? And after the execution of exec command, does shell script process become the some_shell_command process?
Thanks in advance.
update:
The script is:
exec /mnt/usr/bin/pppd $DIAL_DEV unit $count call $PROVIDER ipparam $PROVIDER &
and at sometime:
# Shutdown ppp connection.
pppOff() {
# Get device index.
local index=$1
# Check connection.
pppCheck $index
if [ $? -ne 0 ]; then
echo "invalid pppd: "$index
return 1
fi
# Get pid.
local PID=$PPPD_PID
echo "pppd pid for "$index": "$PID
# Kill
kill -TERM ${PID}
return 0
}
after executing the pppOff, the script itself is killed. So pppd is executed as the same process as the script maybe.
Is some_shell_command executed in another subprocess?
Yes.
after the execution of exec command, does shell script process become the some_shell_command process?
There are two processes, the one spawned for the background becomes some_shell_command. Parent continues execution.
does it mean 'exec' is meaningless?
It has very little meaning in this specific context. Generally, you should expect that Bash optimizes and if Bash finds out there is only one command, it will optimize it to an exec.
$ strace -ff bash -c '/bin/echo 1' 2>&1 | grep clone
# nothing, because `fork()` is optimized out
There are cases (see https://github.com/bminor/bash/blob/f3a35a2d601a55f337f8ca02a541f8c033682247/builtins/evalstring.c#L441 https://github.com/bminor/bash/blob/f3a35a2d601a55f337f8ca02a541f8c033682247/builtins/evalstring.c#L124 ) where the command is not optimized, mostly in the case of like trap or some signal handling that Bash needs to execute after the command is done.
Another difference is that exec requires specifically an executable, where without exec then some_shell_command could be a built-in, function or an alias.

How to execute a command in a different shell?

I have a bash script that opens up a shell called salome shell and it should execute a command called as_run in that shell. The thing is that after entering the salome shell it doesn't execute the command until I exit the salome shell. This is the code that i got:
#!/bin/bash
cd /opt/salome/appli_V2018.0.1_public
./salome shell
eval "as_run /home/students/gbroilo/Desktop/Script/Template_1_2/exportSalome"
What should I do in order to execute the command in the salome shell?
Might be this is what you want:
# call salome shell with commands in a specified script file
cd /opt/salome/appli_V2018.0.1_public
./salome shell <"/home/students/gbroilo/Desktop/Script/Template_1_2/exportSalome"
Or might be this is what you want:
# pipe a command as_run... to salome shell
cd /opt/salome/appli_V2018.0.1_public
echo "as_run /home/students/gbroilo/Desktop/Script/Template_1_2/exportSalome" | ./salome shell
Anyway, you have to read the salome guide about how salome shell call it's script.
Most shells implement a way to pass the commands as parameters, e.g.
dash -c 'x=1 ; echo $x'
You'll need to consult your shell's manual to see if it's possible.
You can also try sending the commands to the standard input of the shell:
echo 'set x = 1 ; echo $x' | tcsh
Using a HERE doc might be a bit more readable in case of complex commands:
tcsh << 'TCSH'
set x = 1
echo $x
TCSH

Entering cshell from bash

I have a bash script and need to run some commands in cshell inside it.
#!/bin/bash
echo entering_to_cshell
csh
echo in_cshell
exit
echo exited_from_cshell
Why doesn't this script run as expected? It only prints entering_to_cshell and it doesn't exit from cshell.
By using
csh
you start a new subshell where your script isn't executed. That's why none of your following commands are executed. Your script waits for this subshell to end which, as you noted, never happens.
Try
csh -c "echo in_cshell"
This way you don't create a new subshell which isn't impacted by your script.
By simply calling csh in your script, you're starting an interactive csh subshell. You'll notice that once you quit from the csh session, your script will then continue to with the subsequent echo and quiting on exit.
To pass a series of commands to csh from you bash script, one approach would be to use the Here Document syntax to redirect commands to csh.
#!/bin/bash
echo entering_to_cshell
csh <<EOF
echo in_cshell
exit
EOF
echo exited_from_cshell
The lines between the EOF entries will be treated as a script that is interpreted by csh.

How to change shells in script

The default shell on the the system is csh but I want to write a script in bash. How do I write a script that will run bash and then convert back to csh at the end.
I tried this but it doesn't work:
bash
var=Hello
echo $var
csh
The command you are looking for is exit. When typing at the keyboard use exit instead of csh to get back to csh. When you enteredcsh, that just started a new csh session on top of the csh and bash sessions already running.
%bash
$ var=Hello
$ echo $var
Hello
$ exit
exit
%
As others have said, when using a script:
#! /bin/bash
var=Hello
echo $var
exit # You don't need exit; but it's okay here.
You don't need to change shells back again. When the script is run, it will be run by a sub-shell (which exits at the end of the script), and the parent shell is unchanged. So, as already suggested, the only thing you have to do is ensure the script is run by the correct shell, and the 'shebang' is the way to do that:
#!/bin/bash
var=Hello
echo $var
That's all it takes.
Define it using the sha bang
#!/bin/bash
at the starting of your file.

How to change argv0 in bash so command shows up with different name in ps?

In a C program I can write argv[0] and the new name shows up in a ps listing.
How can I do this in bash?
You can do it when running a new program via exec -a <newname>.
Just for the record, even though it does not exactly answer the original poster's question, this is something trivial to do with zsh:
ARGV0=emacs nethack
I've had a chance to go through the source for bash and it does not look like there is any support for writing to argv[0].
I'm assuming you've got a shell script that you wish to execute such that the script process itself has a new argv[0]. For example (I've only tested this in bash, so i'm using that, but this may work elsewhere).
#!/bin/bash
echo "process $$ here, first arg was $1"
ps -p $$
The output will be something like this:
$ ./script arg1
process 70637 here, first arg was arg1
PID TTY TIME CMD
70637 ttys003 0:00.00 /bin/bash ./script arg1
So ps shows the shell, /bin/bash in this case. Now try your interactive shell's exec -a, but in a subshell so you don't blow away the interactive shell:
$ (exec -a MyScript ./script arg1)
process 70936 here, first arg was arg1
PID TTY TIME CMD
70936 ttys008 0:00.00 /bin/bash /path/to/script arg1
Woops, still showing /bin/bash. what happened? The exec -a probably did set argv[0], but then a new instance of bash started because the operating system read #!/bin/bash at the top of your script. Ok, what if we perform the exec'ing inside the script somehow? First, we need some way of detecting whether this is the "first" execution of the script, or the second, execed instance, otherwise the second instance will exec again, and on and on in an infinite loop. Next, we need the executable to not be a file with a #!/bin/bash line at the top, to prevent the OS from changing our desired argv[0]. Here's my attempt:
$ cat ./script
#!/bin/bash
__second_instance="__second_instance_$$"
[[ -z ${!__second_instance} ]] && {
declare -x "__second_instance_$$=true"
exec -a MyScript "$SHELL" "$0" "$#"
}
echo "process $$ here, first arg was $1"
ps -p $$
Thanks to this answer, I first test for the environment variable __second_instance_$$, based on the PID (which does not change through exec) so that it won't collide with other scripts using this technique. If it's empty, I assume this is the first instance, and I export that environment variable, then exec. But, importantly, I do not exec this script, but I exec the shell binary directly, with this script ($0) as an argument, passing along all the other arguments as well ($#). The environment variable is a bit of a hack.
Now the output is this:
$ ./script arg1
process 71143 here, first arg was arg1
PID TTY TIME CMD
71143 ttys008 0:00.01 MyScript ./script arg1
That's almost there. The argv[0] is MyScript like I want, but there's that extra arg ./script in there which is a consequence of executing the shell directly (rather than via the OS's #! processing). Unfortunately, I don't know how to get any better than this.
Update for Bash 5.0
Looks like Bash 5.0 adds support for writing to special variable BASH_ARGV0, so this should become far simpler to accomplish.
(see release announcement)
( exec -a foo bash -c 'echo $0' )
ps and others inspect two things, none of which is argv0: /proc/PID/comm (for the "process name") and /proc/PID/cmdline (for the command-line). Assigning to argv0 will not change what ps shows in the CMD column, but it will change what the process usually sees as its own name (in output messages, for example).
To change the CMD column, write to /proc/PID/comm:
echo -n mynewname >/proc/$$/comm; ps
You cannot write to or modify /proc/PID/cmdline in any way.
Process can set their own "title" by writing to the memory area in which argv & envp are located (note that this is different than setting BASH_ARGV0). This has the side effect of changing /proc/PID/cmdline as well, which is what some daemons do in order to prettify (hide?) their command lines. libbsd's setproctitle() does exactly that, but you cannot do that in Bash without support of external tools.
I will just add that this must be possible at runtime, at least in some environments. Assigning $0 in perl on linux does change what shows up in ps. I do not know how that is implemented, however. If I can find out, i'll update this.
edit:
Based on how perl does it, it is non-trivial. I doubt there is any bask built in way at runtime but don't know for sure. You can see how perl does sets the process name at runtime.
Copy the bash executable to a different name.
You can do this in the script itself...
cp /bin/bash ./new-name
PATH=$PATH:.
exec new-name $0
If you are trying to pretend you are not a shell script you can rename the script itself to something cool or even " " (a single space) so
exec new-name " "
Will execute bash your script and appears in the ps list as just new-name.
OK so calling a script " " is a very bad idea :)
Basically, to change the name
bash script
rename bash and rename the script.
If you are worried, as Mr McDoom. apparently is, about copying a binary to a new name (which is entirely safe) you could also create a symlink
ln -s /bin/bash ./MyFunkyName
./MyFunkyName
This way, the symlink is what appears in the ps list. (again use PATH=$PATH:. if you dont want the ./)

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