How do I get $(/bin/printf -6) to return -6 and not think -6 is an option.

Question

I have a bash shell script which has the line:

g=$(/bin/printf ${i})

when ${i} contains something like -6, printf thinks its being passed an option. It does not recognize the option so produces an error.

if wrap ${i} in quotes, printf still thinks its being passed an option.

g=$(/bin/printf "${i}")

if I escape the quotes, variable $g then holds "-6" which is not what I want either.

g=$(/bin/printf \"${i}\")

Is there away to escape the dash (-).

printf is a BusyBox app

Answer 1

Most GNU programs support using -- as a delimiter to tell the program that all further arguments are not options. For instance

$ printf -- -6
-6

Answer 2

You should use

printf -- -6

Answer 3

What if you called printf with an actual format string?

$ printf "%d\n" -6
-6
$ /sbin/busybox printf "%d\n" -6
-6
$

This works with both GNU coreutils' and busybox' printf, apparently.

Answer 4

If you feed a non-numeric argument in this way, you get an error message:

$ busybox printf "%d" "a"
a: conversion error
-1

But you can use %s and it will work for both numeric and non-numeric arguments (as long as you don't need to do any formatting):

$ busybox printf "%s" "a"
a
$ busybox printf "%s" -6
-6

If you're not using the formatting features of printf and you need to output the value without a newline, busybox's echo command supports -n:

$ busybox echo -n "a"
a
$ busybox echo -n -6
-6