Shell function that consumes two arguments per loop iteration

Question

I'm writing some iptables scripts, and I want to write a function that takes an arbitrary number of parameters and consumes them two at a time. Here's an example:

#!/bin/sh
# Allow inbound sessions for a specific service
iptables --append INPUT --protocol $PROTO --destination-port $PORT \
   --match state --state NEW --jump ACCEPT || exit 1

I found this thread that shows the right syntax for looping through an arbitrary number of arguments, but I don't know how to grab two arguments in each iteration. How do I get both $PROTO and $PORT from the caller (from $@, two args at a time)?

I don't quite understand what you're asking. Where do $PROTO and $PORT come from? Where's your arbitrary number of parameters? What is an example of how you'd like this script to be called? — Celada
– Celada, Commented Feb 22, 2017 at 20:53

Stéphane Chazelas · Accepted Answer · 2017-02-22 21:11:38Z

You could do:

#! /bin/sh -
while [ "$#" -ge 2 ]; do
  proto=$1 port=$2
  shift 2
  iptables --append INPUT --protocol "$proto" --destination-port "$port" \
    --match state --state NEW --jump ACCEPT || exit 1
done

With zsh:

#! /bin/zsh -
for proto port do
  iptables --append INPUT --protocol "$proto" --destination-port "$port" \
    --match state --state NEW --jump ACCEPT || exit 1
done

One difference is that if there's an odd number of argument, there will be an extra run with $proto containing the last argument and $port being set but empty (as if we had used [ "$#" -gt 0 ] instead of [ "$#" -ge 2 ] in the previous example).

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