How do I echo "-e"?

Question

I want to echo a string that might contain the same parameters as echo. How can I do it without modifying the string?

For instance:

$ var="-e something"
$ echo $var
something

... didn't print -e

Answer 1

A surprisingly deep question. Since you tagged bash, I'll assume you mean bash's internal echo command, though the GNU coreutils' standalone echo command probably works similarly enough.

The gist of it is: if you really need to use echo (which would be surprising, but that's the way the question is written by now), it all depends on what exactly your string can contain.

The easy case: -e plus non-empty string

In that case, all you need to do is quote the variable before passing it to echo.

$ var="-e something"
$ echo "$var"
-e something

If the string isn't eaxctly an echo option or combination, which includes any non-option suffix, it won't be recognized as such by echo and will be printed out.

Harder: string can be -e only

If your case can reduce to just "-e", it gets trickier. One way to do it would be:

$ echo -e '\055e'
-e

(escaping the dash so it doesn't get interpreted as an option but as on octal sequence)

That's rewriting the string. It can be done automatically and non-destructively, so it feels acceptable:

$ var="-e something"
$ echo -e ${var/#-/\\055}
-e something

You noticed I'm actually using the -e option to interpret an octal sequence, so it won't work if you intended to echo -E. It will work for other options, though.

The right way

Seriously, you're not restricted to echo, are you?

printf '%s\n' "$var"

Answer 2

The proper bash way is to use printf:

printf "%s\n" "$var"

By the way, your echo didn't work because when you run:

var="-e something"
echo $var

(without quoting $var), echo will see two arguments: -e and something. Because when echo meets -e as its first argument, it considers it's an option (this is also true for -n and -E), and so processes it as such. If you had quoted var, as shown in other answers, it would have worked.