How to loop over shell arguments in reverse order?

Question

I want to build a loop in my shell script where i run over all arguments in reverse order and use them in a command. How can this be done?

My specific code would be:

#!/bin/sh
for MODULE in "$@"
do
rmmod $MODULE
done
for MODULE in "$@"
do
insmod $MODULE
done
dmesg -c

but i want the "insmod" in the second for-loop to be called with the arguments from last to first instead.

Do you need to use #!/bin/sh, or are you assuming that you'll be using bash? — chepner
– chepner, Commented Apr 30, 2020 at 16:12
The simplest solution is to build an array in reverse order while traversing "$@", or to use a C-style for loop, but /bin/sh doesn't necessarily support either. — chepner
– chepner, Commented Apr 30, 2020 at 16:13
How can i check what i have available? If I do echo $SHELL on the device i want to run this, i get /bin/sh as answer — heslant
– heslant, Commented Apr 30, 2020 at 16:51
If /bin/bash exists you can (and should) use it in scripts by changing the shebang line to #!/bin/bash. Then you get access to many more features. — John Kugelman
– John Kugelman, Commented Apr 30, 2020 at 17:07

John Kugelman · Accepted Answer · 2020-04-30 16:21:31Z

3

If you're targeting plain sh and not bash then you won't have any array features to speak of. You could use a recursive function:

#!/bin/sh

resetmods() {
    rmmod "$1"
    [ "$#" -gt 1 ] && (shift; resetmods "$@")
    insmod "$1"
}

resetmods "$@"
dmesg -c

answered Apr 30, 2020 at 16:21

John Kugelman

365k70 gold badges555 silver badges600 bronze badges

Sign up to request clarification or add additional context in comments.

1 Answer 1