Escaping strings in Bash using !:q

TIL this trick, [via Pascal Hirsch](https://twitter.com/phphys/status/1311727268398465029) on Twitter. Enter a line of Bash starting with a `#` comment, then run `!:q` on the next line to see what that would be with proper Bash escaping applied. ``` bash-3.2$ # This string 'has single' "and double" quotes and a $ bash-3.2$ !:q '# This string '\''has single'\'' "and double" quotes and a $' bash: # This string 'has single' "and double" quotes and a $: command not found ``` How does this work? [James Coglan explains](https://twitter.com/mountain_ghosts/status/1311767073933099010): > The `!` character begins a history expansion; `!string` produces the last command beginning with `string`, and `:q` is a modifier that quotes the result; so I'm guessing this is equivalent to `!string` where `string` is `""`, so it produces the most recent command, just like `!!` does