bash - How to wrap another shell still passing $OPTIND as-is? -

i'm trying wrap bash script b script a. want pass options passed b are.

#!/bin/bash  # script ./b ${@:$optind} 

this print $1 (if any). what's simplest way not to?

so calling:

./a -c -d 5 first-arg 

i want b execute: ./b -c -d 5 # without first-arg

in bash, can build array containing options, , use array call auxiliary program.

call_b () {   typeset -i i=0   typeset -a a; a=()   while ((++i <= optind));   # i=1..$optind     a+=("${!i}")                # append parameter $i $a   done   ./b "${a[@]}" } call_b "$@" 

in posix shell (ash, bash, ksh, zsh under sh or ksh emulation, …), can build list "$1" "$2" … , use eval set different positional parameters.

call_b () {   i=1   while [ $i -le $optind ];     a="$a \"\$$i\""     i=$(($i+1))   done   eval set -- $a   ./b "$@" } call_b "$@" 

as often, rather easier in zsh.

./b "${(@)@[1,$optind]}"