mjhjmtu

By convention -- signals that there is no more options after it.
It seems to me that when using getopts with case clause, -) pattern subclause doesn't match --. So what is the behavior of getopts when it meets --? Does it treat -- as an option, a nonoption argument, or neither? Thanks.

The behaviour is that it stops parsing the command line and leaves the rest of the arguments as is. The -- itself is removed (or rather $OPTIND will indicate that it was processed but $opt in the code below will never be -, and if you shift "$(( OPTIND - 1 ))" as one usually does, you'll never see it).

Example:

#!/bin/bash

while getopts 'a:b:' opt; do
 case "$opt" in
 a) printf 'Got a: "%s"n' "$OPTARG" ;;
 b) printf 'Got b: "%s"n' "$OPTARG" ;;
 *) echo 'error' >&2
 exit 1
 esac
done

shift "$(( OPTIND - 1 ))"

printf 'Other argument: "%s"n' "$@"

Running it:

$ bash script.sh -a hello -- -b world
Got a: "hello"
Other argument: "-b"
Other argument: "world"

As you can see, the -b world bit of the command line was not processed by getopts.

It stops parsing the command line at -- or at the first non-option argument:

$ bash script.sh something -a hello -- -b world
Other argument: "something"
Other argument: "-a"
Other argument: "hello"
Other argument: "--"
Other argument: "-b"
Other argument: "world"

In this case, -- was not "removed" since getopts never got that far.