How to force decimal interpretation of printf's parameter in Bash instead of octal

Question

I am formatting numbers in a bash script with printf

printf "%04d" $input   # four digits width with leading zeros if shorter

If input="011" printf assumes the value is octal and returns 0009 instead of 0011.

Is it possible to force printf to assume all values are decimal?

Why not just remove the leading zeroes when you assign to `input`? — Barmar, Jan 06 '17 at 19:06
In that case I have to check any input if it starts with a zero. I think the answer from chepner/jackman is more concise. — lojoe, Jan 06 '17 at 19:23
You have to check your input anyway, right? you're not just using user input without checking anything, right? so why not use a regex like `if [[ $input =~ ^0*([[:digit:]]*)$ ]]; then input=${BASH_REMATCH[1]}; printf '%04d\n' "$input"; else echo 'bad input'; fi`. — gniourf_gniourf, Jan 07 '17 at 11:04
@Barmar gave me the idea for `while echo "$input"|grep -q '^0';do input=$(echo "$input"|cut -c2-);done` — Alexx Roche, Dec 30 '18 at 12:54

score 6 · Accepted Answer · edited Jan 06 '17 at 19:05

6

Prefix the value with a base specifier of 10#.

$ printf "%04d\n" "$input"
0009
$ printf "%04d\n" "$((10#$input))"
0011

edited Jan 06 '17 at 19:05

glenn jackman

answered Jan 06 '17 at 19:04

chepner

1

Need to use arithmetic evaluation there – glenn jackman Jan 06 '17 at 19:06
Thanks; I overlooked the error message I got, and `printf` produces output for `10` (after dropping the `#011`) anyway. – chepner Jan 06 '17 at 19:13

1 Answers1