Bash, arithmetic expansion and formatting.

[zandor]

Starting here with aritmaetic expansion I found a problem where I would like to keep leading zeros.

For example I have,

COUNT=$((1))

for ((X=0; X<20; X++))
do
echo Help$((COUNT))
((COUNT++))
done

Will print out:

Help1
Help2

...
Help18
Help19

Is there an easy way to have leading zeros in the variable expansion so the output is something like:

Help0001
...
Help0018
Help0019

I've tried assigning COUNT as a string first, COUNT=0001, but the end result is only the first iteration has the correct amount of digits.

-Zandor

[zandor]

I spoke to soon but I found a solution which is printf. In case anyone else is wondering here's how it works.

printf "%d" 4 will simply print a decimal
4

printf "%4d" 4 will print the same but will align the right side of the number to position 4
---4 (replace the dashes with spaces)
and

printf "%04d" 4 tells printf to fill in the left of the number with leading zeros.
0004

I know it's not much of an explanation but printf is a complex function as it follows the C language formatting arguments.

do a man printf for the bash printf and a man 3 printf for the C manpage.

-Zandor