How do I use the printf command with a format string beginning with "-"? [duplicate]

Question

I have some items:

items=(one two three)

And I want to print them with leading -, to give this output:

- one - two - three

The obvious format string fails:

printf '- %s\n' "${items[@]}"

bash: printf: - : invalid option printf: usage: printf [-v var] format [arguments]

I tried using \- in the format string, but that results in literal \- in the output.

If I could assume ASCII, I might get away with \055, but is there a portable way to write this?

Have a look at Which characters need to be escaped when using Bash? — F. Hauri - Give Up GitHub
– F. Hauri - Give Up GitHub, Commented Mar 17 at 15:01
@F.Hauri, how is that relevant? This argument is just an opaque string as far as the shell itself is concerned - it's single-quoted. — Toby Speight
– Toby Speight, Commented Mar 17 at 15:21

phuclv · Accepted Answer · 2025-03-17 15:11:47Z

4

In most GNU tools just use -- to notify that none of the following parameters are program arguments

$ printf -- '- %s\n' "${items[@]}" - one - two - three

or escape the - character code

$ printf '\055 %s\n' "${items[@]}" $ printf '\x2D %s\n' "${items[@]}"

edited Mar 17 at 15:11

answered Mar 17 at 15:01

phuclv

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3 Comments

Toby Speight Mar 17 at 15:18

I don't know why I forgot --!

phuclv Mar 17 at 15:24

you can also use various other things like %c 0x2D but not as straightforward as the above solutions in this case

Toby Speight Mar 17 at 15:39

that has two problems - code-points aren't portable, and we'd somehow need to get the - in front of every argument. I guess we could insert it into the array expansion instead: printf '%s\n' "${items[@]/#/- }".

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