How to use eval to read variables

Question

Here is the scenario: suppose I set positional variables:

set 1 2 3 4
eval "args_$1=something"

how do I read args_1,args_2,args_3... variables I tried

echo $args_$1

and this also not working

eval "\$$(echo arg_$1)"

How do I get value of $arg_1, to display on terminal or pass to a function, etc.

Using eval is not recommended but, you can try: set 1 2 3 4; eval arg_$1=koba; eval echo $\`echo arg_$1\` — AKS, Apr 14 '15 at 00:44
The answer I most strongly endorse from the linked question, by the way, is http://stackoverflow.com/a/16973754/14122 -- which, in the case given here, would mean something like `printf -v "$args_$1" %s value` — Charles Duffy, Apr 14 '15 at 01:38
...but really, read BashFAQ #6 for a more complete answer / set of approaches than what's available here on StackOverflow. — Charles Duffy, Apr 14 '15 at 01:45

score 2 · Answer 1 · answered Apr 14 '15 at 01:31

2

Without eval:

$ set 1 2 3 4
$ var="args_$1"
$ declare "$var=foo"
$ echo "$var"
args_1
$ echo "${!var}"
foo

answered Apr 14 '15 at 01:31

glenn jackman

score 1 · Accepted Answer · answered Apr 14 '15 at 00:46

1

$ set 1 2 3 4; eval arg_$1=koba; eval echo $`echo arg_$1`
koba

PS: Using eval is not recommended.

answered Apr 14 '15 at 00:46

AKS

2 Answers2