initializing arrays in sh

Question

I want to initialize an array in sh.

In bash that would be:

list=(`seq 1 4`)

In sh I try to do it like this:

    for i in `seq 1 4`; do
        list[$((i-1))]="$i"
    done

I get an error though for each iteration saying:

list[0]=1: not found

What am I doing wrong and how to fix that?

This is more or less a duplicate of http://stackoverflow.com/questions/6499486/how-to-mark-an-array-in-posix-sh — Adam Spiers, Dec 08 '13 at 16:39

score 4 · Accepted Answer · answered Dec 08 '13 at 16:35

4

POSIX sh doesn't support arrays. You need a more advanced shell for that, e.g. bash, zsh, or ksh.

answered Dec 08 '13 at 16:35

Adam Spiers

17,397
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1

Not really. It just fails to mention that only some shells support it. – Adam Spiers Dec 08 '13 at 16:38
`sh on OSX` seems to support arrays. – anubhava Dec 08 '13 at 16:52
1

Only because `sh` on OS X is really `bash`, which is a bit loose about what it accepts in "POSIX" mode. – chepner Dec 08 '13 at 17:03

score 1 · Answer 2 · answered Dec 08 '13 at 16:50

1

If you really want to use arrays, you can fudge them by writing your own array function. I'm not going to encourage this by giving you a full function :-) but here's the gist:

$ f0=yay 
$ t=0
$ eval echo f$t
f0
$ eval echo \$f$t
yay

answered Dec 08 '13 at 16:50

rand'Chris

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initializing arrays in sh

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