How to represent hexadecimal number with two digits in linux scripts (bash)

Question

I have a problem. I must represent each number with exactly 2 digits (i.e. 0F instead of F)

my code looks something like that:
1. read number of ascii chars are in an environment variable (using wc -w)
2. converting this number to hexadecimal (using bc).

How can I make sure the number I'll get after stage 2 will include a leading 0 (if necessary)

Thanks, Amigal

Answer 1

Run it through printf:

$ printf "%02s" 0xf
0f

If you have the digits of the number only in a variable, say $NUMBER, you can concatenate it with the required 0x prefix directly:

$ NUMBER=fea
$ printf "%04x" 0x$NUMBER
0fea

Answer 2

Skip the conversion part in bc and convert the decimal number with the printf command:

#num=$(... | wc -w)
num=12  # for testing
printf "%02x" $num
0c

If you must convert to hexadecimal for other reasons, then use the 0x prefix to tell printf that the number is already hexadecimal:

#num=$(... | wc -w | bc ...)
num=c   # as obtained from bc
printf "%02x" 0x$num
0c

The %x format requests formatting the number as hexadecimal, %2x requests padding to width of two characters, and %02x specifically requests padding with leading zeros, which is what you want. Refer to the printf manual (man printf or info coreutils printf) for more details on possible format strings.