How to printf an exclamation mark in bash? [duplicate]

Question

I need to printf a simple script and redirect the output to a file, but when I do this:

printf "#!/bin/bash\ntouch /tmp/1234567890_$RUN" > /tmp/password-change-script_$RUN.sh

I get this error:

bash: !/bin/bash\ntouch: event not found

If I escape the exclamation mark:

printf "#\!/bin/bash\ntouch /tmp/1234567890_$RUN" > /tmp/password-change-script_$RUN.sh

Then the escape character is still present in the file.

cat /tmp/password-change-script_$RUN.sh
#\!/bin/bash
touch /tmp/1234567890_111

By the way, in this particular case, the #!/bin/bash MUST be in the file. For some reason the binary file that executes the script won't read the file otherwise.

Answer 1

The ! character is expanded in double-quoted strings, but not in single-quoted strings.

printf '#!/bin/bash\ntouch /tmp/1234567890_'"$RUN"

It's also not expanded when it appears by itself or at the end of a word; this isn't as clean but:

printf "#%c/bin/bash\ntouch /tmp/1234567890_$RUN" !

You can also temporarily turn off history substitution by (temporarily) setting $histchars to the empty string; this turns off the special treatment of !:

histchars=
printf "#!/bin/bash\ntouch /tmp/1234567890_$RUN"
unset histchars

Or you can execute the printf command in a script rather than interactively (history substitution is on by default only for interactive shells).