Here is a test:
$ bash -c "pgrep -f novalidname"
$ sh -c "pgrep -f novalidname"
11202
Why is pgrep giving output when run from sh? (As far as I can see, there are no processes on my computer that is named novalidname)
It's probably a timing issue and pgrep finds itself, as you're issuing it with -f and novalidname is present in the command line. Try with -l to confirm.
The actual explanation:
Regardless of flags, pgrep never returns its own PID.
If you execute bash -c with a simple command, then bash will exec the command rather than creating a redundant subshell to execute it in. Consequently, bash -c "pgrep -f blah" will replace the bash process with a pgrep process. If that pgrep process is the only process whose command line includes blah, then pgrep will not display any PIDs (as per 1).
dash does not perform the above optimization. (zsh and ksh do.) So if on your system, sh is implemented with dash, then sh -c "pgrep -f blah" will result in two processes being executed -- the sh process and the pgrep child -- both of which contain blah in their command lines. pgrep will not report itself, but it will report its parent.
That's one thing (finding itself because of delay) see also:
$ ps ax | grep novalidname
Here it usually shows as well. (on Ubuntu does for me. (under bash)
The other thing is what is /bin/sh bound to?
On most Linux distros /bin/sh is a soft link to default shell which is usually actually bash, but can be any other shell.
The time difference that causes grep/pgrep to show itself may be introduced by finding a soft link location (hm, odd) or some other shell is bound to /bin/sh which executes slightly different than bash, thus causing the delay needed for process to show in pgrep.
Also, bash will firstly try to source ~/.bashrc and load its history, while /bin/sh will do what will do. In .bashrc can be pgrep defined as alias in another way which may also affect the difference.
To see where /bin/sh points to do:
$ readlink -e /bin/sh
Or just run sh to see what will show up. :D
Related
I want to check if a command exists on my machine (RedHat) inside a perl script.
Im trying to check if compgen -c contains the desired command, but running it from inside a script just gives me an empty output. Other commands work fine.
example.pl:
my $x = `compgen -c`;
print $x;
# empty output
my $y = `ls -a`;
print $y;
# .
# ..
# example.pl
Are there possible solutions for this? Or is there a better way to check for commands on my machine?
First, Perl runs external commands using /bin/sh, which is nowadays a link to a shell that is a default-of-sorts on your system. Much of the time that is bash, but not always; on RedHat it is.
This compgen is a bash builtin. One way to discover that is to run man compgen (in bash) -- and the bash manual pops up. Another way is type as Dave shows.
To use builtins we generally need to run an explicit shell for them, and they have a varied behavior in regards to whether the shell is "interactive" or not.† I can't find a discussion of that in bash documentation for this builtin but experimentation reveals that you need
my #completions = qx(bash -c "compgen -c")
The quotes are needed so to pass a complete command to a shell that will be started.
Note that this way you don't catch any STDERR out of those commands. That will come out on the terminal, and it can get missed that way. Or, you can redirect that stream in the command, by adding 2>&1 (redirect to STDOUT) at the end of it.
This is one of the reasons to use one of a number of good libraries for running and managing external commands instead of the builtin "backticks" (the qx I use above is an operator form of it.)
† This can be facilitated with -i
my #output_lines = qx(bash -i -c "command with arguments")
It's because compgen is a bash built-in command, not an external command. And when you run a command using backticks, you get your system's default shell - which is probably going to be /bin/sh, not bash.
The solution is to explicitly run bash, using the -c command-line option to give it a command to run.
my $x = `bash -c compgen -c`;
From a bash prompt, you can use type to see how a command is implemented.
$ type ssh
ssh is /usr/bin/ssh
$ type compgen
compgen is a shell builtin
Perl 6's shell sends commands to the "shell" but doesn't say what that is. I consistently get bash on my machine but I don't know if I can rely on that.
$ perl6 -e 'shell( Q/echo $SHELL/ )'
/bin/bash
$ csh
% perl6 -e 'shell( Q/echo $SHELL/ )'
/bin/bash
% zsh
$ perl6 -e 'shell( Q/echo $SHELL/ )'
/bin/bash
That's easy enough on Unix when it's documented, but what about cmd.exe or PowerShell on Windows (or bash if it's installed)? I figure it's the cmd.exe but a documented answer would be nice.
Looking at the source, rakudo just calls /bin/sh -c on non-windows and uses %*ENV<ComSpec> /c on windows.
dash (installed as /bin/sh on many systems), doesn't set $SHELL, nor should it. $SHELL isn't the name of the parent process; it's the name of the shell that should be used when an interactive shell is desired.
To get the name of the parent process, you could use the following on some systems:
echo "$0"
or
# Command line
perl -e'$ppid=getppid(); #ARGV="/proc/$ppid/cmdline"; CORE::say "".<>'
or
# Program file
perl -e'$ppid=getppid(); CORE::say readlink("/proc/$ppid/exe")'
You'll find you'll get /bin/sh in all cases.
I'm wondering if there a way of capturing a list of the processes executed on a non-interactive shell?
Basically I have a script which calls some variables from other sources and I want to see what the values of said variables are. However, the script executes and finishes very quickly so I can't capture the values using ps.
Is there a way to log processes and what arguments were used?
TIA
Huskie
EDIT:
I'm using Solaris in this instance. I even thought about about having a quick looping script to capture the values being passed - but this doesn't seem very accurate and I'm sure executions aren't always being captured.
I tried this:
#!/bin/ksh
while [ true ]
do
ps -ef | grep $SCRIPT_NAME |egrep -v 'shl|lis|grep' >> grep_out.txt
done
I'd use sleep but I can't specify any precision as all my sleep executables want an integer value rather than any fractional value.
On Solaris:
truss -s!all -daDf -t exec yourCommand 2>&1 | grep -v ENOENT
On AIX and possibly other System V based OSes:
truss -s!all -daDf -t execve yourCommand 2>&1 | grep -v ENOENT
On Linux and other OSes supporting strace, you can use this command:
strace -ff -etrace=execve yourCommand 2>&1 >/dev/tty | grep -v ENOENT
In case the command you want to trace is already running, you can replace yourCommand by -p pid with pid being the process to be traced process id.
EDIT:
Here is a way to trace your running script(s) under Solaris:
for pid in $(pgrep -f $SCRIPT_NAME); do
truss -s!all -daDf -t exec -p $pid 2>&1 | grep -v ENOENT > log.$pid.out &
done
Note that with Solaris, you might also use dtrace to get the same (and more).
Most shells can be invoked in debug mode, where each statement being executed is printed to stdout (or stderr) after variable substitution and expansion.
For Bourne like shells (sh, bash), debug is enabled with the -x option (as in bash -x myscript) or using the set -x statement within the script itself.
However, debugging only works for the 'current' script. If the script calls other scripts, these other scripts will not execute in debug mode. Furthermore, the code inside functions may not be executed in debug mode either - depends on the specific shell - although you can use set -x within a function to enable debug explicitly.
A very much more verbose (at least by default) option is to use something like strace for this.
strace -f -o trace.out script.sh
will give you huge amounts of information about what the script is doing. For your specific usage you will likely want to limit the output a bit with the -e trace=.... option to control which system calls are traced.
Use truss instead of strace on Solaris. Use dtruss on OS X (I believe). With appropriate command line argument changes as well.
I have a host on which I don't have sudo. Its been setup with ksh, I'm too used to bash and chsh doesn't work. So I put in a /bin/bash as the first line in the .profile on the system.
So the result is, when I login to this system, it automatically gets me into bash. However, when I exit the shell, not suprisingly I land up in ksh.
Any tricks to avoid this?
Use exec to replace the current process (shell) with the new process (shell).
I recommend two steps:
if [ $SHELL != /bin/bash ]
then SHELL=/bin/bash exec /bin/bash --login
fi
Or, you can compress that to:
[ $SHELL != /bin/bash ] && SHELL=/bin/bash exec /bin/bash --login
You can then put the rest of your Bash profile after this. Note that probably you don't put a shebang on the first line - that will confuse things. Also, while testing, make sure you have a second connection (window) open so that you can adjust problems. It is annoying to get locked out by an erroneous profile.
You may write a script named myexit like this:
kill -1 $(ps | sed 1d | awk '{print $1}')
It sends the signal hang up (SIGHUP) to process attached to this terminal.
And would not affect any process started up by nohup.
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 ./)