Echoing an echo option

Question

Today I was testing a script and for some reason I needed to print -n to the console.

I tried echo -n, which of course just echos without a new line. I tried echo "-n", which does the same.

Then I tried assigning to a variable: str="-n". Of course echo $str and echo "$str" do nothing.

How would you echo exactly -n?

@Barmar This won't work, `printf '-n' -bash: printf: -n: invalid option printf: usage: printf [-v var] format [arguments]` — jay, Jan 07 '22 at 21:14
Related: [echo "-n" will not print -n?](/q/13258664/4518341) — wjandrea, Jan 07 '22 at 21:56
Dupe of [StackExchange post](https://unix.stackexchange.com/questions/168095/how-do-i-print-e-with-echo) which has solutions with both `printf` and `echo`. — quickshiftin, Jan 07 '22 at 22:31

score 5 · Answer 1 · answered Jan 07 '22 at 21:14

5

Use printf rather than echo

printf '%s\n' '-n'

answered Jan 07 '22 at 21:14

Barmar

score 0 · Answer 2 · edited May 25 '22 at 09:50

0

You could escape (\) both of those characters, telling your shell to read them both as literal characters.

Running echo \-\n gives me -n as output.

edited May 25 '22 at 09:50

buddemat

answered May 25 '22 at 07:14

BirdBirdBird

2

Escaping it for the shell doesn't prevent `echo` from treating the `-` as an option indicator. And different versions of `echo` behave differently, so what works for your version may not for others -- see [this answer on U&L about `printf` vs `echo`](https://unix.stackexchange.com/questions/65803/why-is-printf-better-than-echo/65819#65819). (BTW, use [backticks to invoke code format](https://meta.stackoverflow.com/questions/251361/how-do-i-format-my-code-blocks).) – Gordon Davisson May 25 '22 at 07:46

2 Answers2