How to mark an array in POSIX sh?

Question

While replacing external commands in a shell script, I used an array to get rid of awk's NF.

Now, since I moved from bash to POSIX sh, I cannot get the array marked right:

#!/bin/bash export RANGE="0 1 4 6 8 16 24 46 53" RANGE=($RANGE) echo arrayelements: $((${#RANGE[@]})) LAST=$((${#RANGE[@]}-1)) echo "Last element(replace NF): ${RANGE[$LAST]}"  # ./foo arrayelements: 9 Last element(replace NF): 53 

I'm using OpenBSD's, sh and it has exactly the same size as the ksh. Changing above to /bin/sh, it seems that the following doesn't work:

set -A "$RANGE" set -- "$RANGE" 

How could I realise the above script in /bin/sh? (Note that it works fine if you invoke bash with --posix, that's not what I look for.)

Answer 1

Arrays are not part of the POSIX sh specification.

There are various other ways to find the last item. A couple of possibilities:

#!/bin/sh export RANGE="0 1 4 6 8 16 24 46 53" for LAST_ITEM in $RANGE; do true; done echo "Last element(replace NF): $LAST_ITEM" 

or:

#!/bin/sh export RANGE="0 1 4 6 8 16 24 46 53" LAST_ITEM="${RANGE##* }" echo "Last element(replace NF): $LAST_ITEM" 

Answer 2

You can use the following project from Github, which implements a POSIX-compliant array, which works in all shells I tried: https://github.com/makefu/array

It is not very convenient to use, but I found it to work well for my purposes.