How to echo #!/ using shell or bash

Question

I am trying to print #!/ via Bash, but it doesn't print it and instead prints the following.

parth@parth-ubuntu64:$ echo "#!\/"
bash: !\/: event not found

EDITED: Allow me to make one more update. How do you make the following work? I think I should just use python instead of bash. The reason why I am trying to do this is because, I want to pipe this into a file which requires sudo permission. This is the only way I can execute sudo => sudo sh -c " echo something > file "

sh -c 'echo \'#!/bin/bash \' '

Re: your edit: `sh -c 'echo "#!/bin/bash"'` will work: `sh` has no history like Bash, and in this line, the `!` is in the single quoted string, so you won't have history expansion. — gniourf_gniourf, Oct 28 '14 at 13:05
You can use `tee` to work around the permission issue: `echo '#!/bin/bash' | sudo tee target-file > /dev/null`. — chepner, Oct 28 '14 at 13:44

Tom Fenech · Accepted Answer · 2014-10-28T12:40:53.707

Within double quotes (and with history expansion enabled), the ! has special meaning. To print those characters literally, use single quotes instead of double:

$ echo '#!\/'
#!\/

History expansion can be disabled in the interactive shell by using set +H, which allows you to use double quotes:

$ set +H
$ echo "#!\/"
#!\/

In general, if you do not require variables to be interpolated (such as in this case), you should use single quotes.

Note that it is not necessary to escape the /, so you can remove the \ from before it if your desired output is #!/:

$ echo '#!/'
#!/

John Kugelman · Answer 2 · 2014-10-28T13:55:31.970

In an interactive shell ! is subject to history expansion. For example, !! is the last command you typed and !$ is the last argument to that command. To avoid history expansion, escape the ! with a backslash or use single quotes.

echo '#!/'

This isn't necessary in shell scripts, only in interactive shells (unless you purposefully enable set -o history and set -o histexpand in the shell script).

Allow me to make one more update. How do you make the following work?
sh -c 'echo \'#!/bin/bash \' '

There's no way to have a single quote inside single quotes. \' doesn't work. So back to double quotes we go.

sh -c "echo '#\!/bin/bash'"

You can also swap the `'` and `"` as in my comment to the OP: `sh -c 'echo "#!/bin/bash"'` — gniourf_gniourf, Oct 28 '14 at 14:06

score 1 · Answer 3 · answered Oct 28 '14 at 12:37

1

Some things are interpreted inside double quotes, including both backslashes and history events introduced by an exclamation mark. Use the more powerful single quotes instead:

echo '#!\/'

answered Oct 28 '14 at 12:37

Peter Westlake

4,894
1
26
35

Actually I understand now what you were trying to say. :) It does work – Parth Oct 28 '14 at 12:39

score 0 · Answer 4 · answered Oct 28 '14 at 12:35

0

You need to put the text inside single quotes, so that it won't parse.

$ echo '#!/'
#!/
$ echo '#!\/'
#!\/

answered Oct 28 '14 at 12:35

Avinash Raj

172,303
28
230
274

score 0 · Answer 5 · edited Oct 28 '14 at 12:43

0

use single quotes instead of double.

echo '#!/'

edited Oct 28 '14 at 12:43

gniourf_gniourf

44,650
9
93
104

answered Oct 28 '14 at 12:35

TrackZero

126
6

How to echo #!/ using shell or bash

5 Answers5