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Full Discussion: Pid=$!
Top Forums Shell Programming and Scripting Pid=$! Post 302843909 by googlietdr on Thursday 15th of August 2013 03:59:52 PM
Old 08-15-2013
Pid=$!

Hello,

I would like to know what this command means?

PID=$!

what does "!" stand for?
 

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PLDD(1) 							 Linux User Manual							   PLDD(1)

NAME
pldd - display dynamic shared objects linked into a process SYNOPSIS
pldd PID pldd OPTION DESCRIPTION
The pldd command displays a list of the dynamic shared objects that are linked into the process with the specified process ID. The list includes the libraries that have been dynamically loaded using dlopen(3). OPTIONS
-?, --help Display program help message. --usage Display a short usage message. -V, --version Display the program version. VERSIONS
pldd is available since glibc 2.15. CONFORMING TO
The pldd command is not specified by POSIX.1. Some other systems have a similar command. EXIT STATUS
On success, pldd exits with the status 0. If the specified process does not exist, the user does not have permission to access its dynamic shared object list, or no command-line arguments are supplied, pldd exists with a status of 1. If given an invalid option, it exits with the status 64. EXAMPLE
$ echo $$ # Display PID of shell 1143 $ pldd $$ # Display DSOs linked into the shell 1143: /usr/bin/bash linux-vdso.so.1 /lib64/libtinfo.so.5 /lib64/libdl.so.2 /lib64/libc.so.6 /lib64/ld-linux-x86-64.so.2 /lib64/libnss_files.so.2 NOTES
The command lsof -p PID also shows output that includes the dynamic shared objects that are linked into a process. SEE ALSO
ldd(1), lsof(1), dlopen(3), ld.so(8) GNU
2014-09-27 PLDD(1)
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