How to echo an exclamation point without the backslash in the output to prevent expansion

Question

I usually like to echo certain string which have an exclamation point in them. And while i know how to escape them to prevent expansion by the Shell, i noticed it also print the backslash...

echo "\!" # will print \! instead of just !

Contrary to other tool like sed which does not print the backslash if used to prevent expansion of possible valid parameter used by sed:

echo "test" | sed 's/^\(.*\)$/\1 \!/' # print -> test !
echo "test" | sed 's/^\(.*\)$/\1 !/' # also print -> test !
echo "test" | sed "s/^\(.*\)$/\1 ! /" # also print -> test !, though needed to put a space or use a backslash for it to show correctly

How can i do the same with echo or I'm i obliged to use other tools like sed?

EDIT:

Just noticed using '' instead of "" with the echo example above works for printing ! without shell expansion (both with and without the backslash), though:

• It wouldn't work in instance when one want to use command substitution with echo, since it only work with ""...

Answer 1

You need to escape it or quote it, not both:

echo \!
echo "!"

Bash history expansion ignores escaped exclamation marks, single-quoted exclamation marks, and exclamation marks which appear just before a closing double quote. See the History Interaction section of the manual for details. So

echo '!'

also works, and without having to think about where the exclamation mark is.

You can combine single quotes with variable expansion by using printf instead of echo. You can also combine multiple forms of quoting in the same command-line:

echo '!'"$(echo test)"

In general though, printf is better than echo. As others have mentioned you can also disable history expansion entirely; non-interactive shells don’t enable it by default.