What is the reason for using shift $ ((OPTIND-1)) after getopts?

My understanding is that shift moves the cli array args n space to the left, and the default n is 1. This means that I can assign values โ€‹โ€‹to the array to existing variables using the $ 1 shift inside the while loop. I don't quite understand why it is used in this context below. The input arguments have already been assigned to values โ€‹โ€‹and removing the $ ((OPTIND-1)) offset does not change this fact. Source: http://linux.die.net/man/3/optind

while getopts ":h:a:fc" opt; do
    case $opt in
            exit 0
            echoerr "Invalid option -$OPTARG"
            exit 1

shift $((OPTIND-1))



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1 answer

Offset removes the parameters processed by the loop getopts

from the parameter list, so that the rest of the script can process the rest of the command line (if any) with $ 1 ... in the usual way, disregarding the number of options processed getopts


Consider a hypothetical script using

frobble [-f] [-c] [-a hostname] filename ...


The getopts

above loop takes care of parsing -f

, -c

and -a

if present, it does not remove them from the argument list. This means that in order to get the filename argument, you need to figure out how many options were processed and continue processing from there. Conveniently, getopts

it tells you the index of the first raw argument: variable OPTIND


And instead of messing up offsets and stuff, you can simply discard the processed parameters by renumbering the rest of the arguments so that your filename is always there $1


This is what it does shift $((OPTIND-1))




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