bash loop and return (10 to 1 and 1 to 10)

i'm currently working on a small piece of script in bash. As a beginner I can't find a solution to a little problem. I need to do a chrono which decreases with each iteration, but when it's equal to 0, it goes back to 10 (10 -> 0 then 0 -> 10). I wrote a small piece that currently doesn't work.

chrono=5
incremant=-1
 while [ $chrono -ge 0 ];
 do
  echo $chrono
  chrono=$((chrono+$incremant))
     if [ $chrono -eq 1 ];
     then
        [ $incremant=1 ];
     fi
 done

I know it's a really simple problem but I'm really stuck at it. Thank you in advance, have a nice day.

Solution

Not sure why you're complicating things with manual loops when bash has perfecly good C-like loops built in:

fn() {
    echo $1
    #printf "%2d * %2d = %3d\n" $chrono $chrono $((chrono * chrono))
}

for ((chrono=10; chrono > 0; chrono--)); do fn $chrono ; done
for ((chrono=1; chrono <= 10; chrono++)); do fn $chrono ; done

You can just change the loop sections depending on what you actually need, the current one counts ten to one inclusive then one to ten inclusive, so duplicates 1. For example, if you don't want the duplicate, just change the first section of the second loop to chrono=2.

You can also perform arbitrarily complex operations on the value with the fn function (such as the commented-out bit which gives you a nicely formatted list of expressions giving the square of each number, like 10 * 10 = 100). The current function just echoes the value: