I have this shell command:
kill `cat -- $PIDFILE`
What the double -- does here? Why not use just
kill `cat $PIDFILE`
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The -- tells cat not to try to parse what comes after it as command line options.
As an example, think of what would happen in the two cases if the variable $PIDFILE was defined as PIDFILE="--version". On my machine, they give the following results:
$ cat $PIDFILE
cat (GNU coreutils) 6.10
Copyright (C) 2008 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.
Written by Torbjorn Granlund and Richard M. Stallman.
$ cat -- $PIDFILE
cat: --version: No such file or directory
Mikael Auno
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3It is worth noting that this behavior (while very common) is defined by the receiving program (i.e. `cat`) and not by the shell. – dmckee --- ex-moderator kitten Feb 21 '10 at 02:33
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Is there any documentation or tutorial on writing your own shell script that understands that `--` means the end of command line options? I've seen ones with getopts and other techniques, but nothing discussing `--`. – CMCDragonkai Aug 25 '15 at 08:08
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3@CMCDragonkai You need not look any further than the [`getopt(1)` man page](http://linux.die.net/man/1/getopt): "Each parameter after a '--' parameter is always interpreted as a non-option parameter". – Mikael Auno Aug 25 '15 at 10:29
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POSIX.1-2017
POSIX also specifies it at: http://pubs.opengroup.org/onlinepubs/9699919799/basedefs/V1_chap12.html#tag_12_02
12.2 Utility Syntax Guidelines
Guideline 10:
The first -- argument that is not an option-argument should be accepted as a delimiter indicating the end of options. Any following arguments should be treated as operands, even if they begin with the '-' character.
Ciro Santilli OurBigBook.com
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