I'm trying to use printf to format some pretty output in a bash script

| This is some output |

But I've stumbled over some behavior I don't understand.

$ printf "--" gives me the error printf: usage: printf [-v var] format [arguments]

and $ printf "-stuff" results in -bash: printf: -s: invalid option

So apparently printf thinks I'm trying to pass some arguments while I'm not.

Meanwhile, completely by accident, I've found this workaround:
$ printf -- "--- this works now ----\n" gives me --- this works now ----

Can anyone explain this behavior?

  • 1
    See also unix.stackexchange.com/questions/11376/… – manatwork Oct 17 '11 at 11:51
  • Out of intereset, are there any implementations of echo that would fail when doing echo ------------? Most only support -n (no trailing newline), -e (interpret backslash-escaped chars) and possible -E (do NOT interpret them) and do not error out when other option-like arguments are encountered, right? (EDIT: GNU's /bin/echo also supports --help and --version.) – janmoesen Oct 18 '11 at 6:34

The -- is used to tell the program that whatever follows should not be interpreted as a command line option to printf.

Edit: Thus the printf "--" you tried basically ended up as "printf with no arguments" and therefore failed.

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-- is being interpreted as an option (in this case, to signify that there are no more options).

A format string should always be included when using printf to prevent bad interpretation. For your particular case:

printf '%s\n' '-----------------------'
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  • what about when you need to do this with a sequence? printf '-%.0s' {1..54} – qodeninja Aug 18 '19 at 18:42
  • 2
    @qodeninja Then you actually do want to use -- :-) printf -- '-%.0s' {1..54} – Chris Down Aug 19 '19 at 14:33

There are differences between printf builtin and /usr/bin/printf, the second one do "what you mean" without these annoying errors.

printf "-----"             # error
printf -- "-----"          # ok
/usr/bin/printf "-----"    # ok
/usr/bin/printf -- "-----" # ok
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