Unix script print arguments

Question

I would like to print a few arguments given in the console
but I would like to print the argument with a number given from a integer

declare -i I=2
declare -i I=4

I would like to print the arguments number 2 and number 4 how can I do that without using the following if statements

if [ $I -eq 2 ]; then
echo $2
fi

What I am searching for is somethink like this
echo $($I) #first access $I, which is 4 and
# then print $4, which is the 4th argument

Some bit of research helps. This has been asked (and answered) scores of times on Stack Overflow. — devnull, May 11 '14 at 11:33

score 1 · Accepted Answer · answered May 11 '14 at 11:29

1

Looks like you're looking for variable indirection. Use like this:

func() {
    p=4
    echo "${!p}"
}

TESTING:

func aa bb cc dd ee
dd

answered May 11 '14 at 11:29

anubhava

761,203
64
569
643

score 1 · Answer 2 · answered May 11 '14 at 11:35

1

To see what each argument is assigned to you can use this loop.

for n in $(seq 1 $#)
do
  echo $n ${!n}
done

answered May 11 '14 at 11:35

John C

4,276
2
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28

Unix script print arguments

2 Answers2