Assign Command line arguments to a variable in for loop in Unix

Question

I have the following bash code:

for (( i=4; i<=$var; ))
do
    temp=`echo $i`
done

I need to assign the command line argument stored in $4 to the variable temp which is not happening.

Not sure what you're asking but it seems like you want temp=$4. — devnull
– devnull, Commented Apr 7, 2014 at 18:13
Prior to adding an echo you should put all your commands in the variable then echo the information — Robert Lee
– Robert Lee, Commented Apr 7, 2014 at 18:16
Why would you use a loop if you only need to work with $4? — Henk Langeveld
– Henk Langeveld, Commented Apr 7, 2014 at 18:20
i have a feeling what he really wants is to loop through all the command line args — CrayonViolent
– CrayonViolent, Commented Apr 7, 2014 at 18:24

that other guy · Accepted Answer · 2014-04-07 18:54:12Z

1

Given a variable i=4 you can get the value of $4 using ${!i}:

set -- foo bar baz thisOne etc
i=4
echo "${!i}"

This prints the fourth positional parameter, thisOne.

answered Apr 7, 2014 at 18:54

that other guy

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