0

How can I use a dash in printf string?

x=xxx
printf -v args "-x=%s" "$x"
printf "$var"

But it gives an error:

./tmp.sh: line 2: printf: -x: invalid option
printf: usage: printf [-v var] format [arguments]

I know that it's because printf interprets -x as an option, but how to overcome it?

I use printf to create a string that will correspond to command line arguments like -x=xxx -y=yyy -z=zzz. Then I want to call a tool like

eval $tool args

  • 2
    You don't want to use printf at all, as you should be building an array of arguments to pass to tool: foo=(-x xxx -y yyy); tool "${foo[@]}" – chepner Jul 25 '18 at 14:02
5

printf supports the conventional end-of-options argument --:

$ printf -- '-x\n'
-x
  • 2
    POSIX requires printf to deal with --. – schily Jul 25 '18 at 14:01
  • 1
    Such support is usually indicated by requiring conformance to the Utility Syntax Guidelines, but printf makes no mention of such a requirement. (At least that I could find; I often fail to find things in the place where I expect them.) – chepner Jul 25 '18 at 14:06
  • POSIX gives no permission to disregard the USG for printf. – schily Jul 25 '18 at 14:18
  • dash is not POSIX compliant in many cases. It would e.g. need to support multi byte characters in case it is not run only on a tiny embedded system. – schily Jul 25 '18 at 14:19
  • 2
    @chepner, see pubs.opengroup.org/onlinepubs/9699919799/utilities/…. Standard utilities that do not accept options, but that do accept operands, shall recognize "--" as a first argument to be discarded. – Stéphane Chazelas Jul 25 '18 at 14:26
3

You could use the end of options (--) argument to have printf treat the dash literally:

printf -v var -- '-x=%s' "$x"

Output:

$ x=xxx && printf -v var -- '-x=%s' "$x" && echo $var
-x=xxx
2

Since your printf seems to take options, use a format string to output both the -x and the value of the variable:

printf '%s=%s' '-x' "$x"

To call a tool with a pre-generated command line, don't put the command line arguments into a string. Use set to set the values of $@ instead:

set -- -x="$x" -y="$y" -z="$z"

Then call the tool:

tool "$@"

This will ensure that the tool gets the correct number of arguments, properly quoted, even if one or several of the values, $x, $y, or $z, contains spaces.

Or, using a bash array:

args=( -x="$x" -y="$y" -z="$z" )

tool "${args[@]}"

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