I have a bash shell script which has the line: <pre class="prettyprint"><code>g=$(/bin/printf ${i}) </code></pre> when <code>${i}</code> contains something like <code>-6</code>, <code>printf</code> thinks its being passed an option. It does not recognize the option so produces an error. if wrap <code>${i}</code> in quotes, <code>printf</code> still thinks its being passed an option. <pre class="prettyprint"><code>g=$(/bin/printf "${i}") </code></pre> if I escape the quotes, variable <code>$g</code> then holds "<code>-6</code>" which is not what I want either. <pre class="prettyprint"><code>g=$(/bin/printf \"${i}\") </code></pre> Is there away to escape the dash (-). printf is a BusyBox app

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

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

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

$ printf -- -6
-6