bash getopts not recognizing the second argument

Question

This is my entire script in its simplest form.

#!/bin/bash
src=""
targ=${PWD}

while getopts "s:t:" opt; do
  case $opt in
    s)
      src=$OPTARG
      ;;
    t)
      targ=$OPTARG
      ;;
  esac
  shift $((OPTIND-1))
done

echo "Source: $src"
echo "Target: $targ"

I run this script as getopts_test -s a -t b

However, it always prints the pwd in front of the Target: and never b

What am I missing here?

Answer 1

The reason for why b is never printed is that the shift within the loop moves the processed options away after the first iteration, i.e. after a has been printed. Use of shift $((OPTIND-1)) is intended to access the possible given variadic parameters. Naturally, once you remove shift, targ gets reassigned to b, and ${PWD} is no longer included in it since you don't have concatenation of the strings (targ and the option of -t) anywhere.