Ethan Gascoigne <esgascoi...@gmail.com> writes:
> Narrowed it down to the line `eval "$(fasd --init auto)"` in my
> .bashrc. Must be some kindof bug with fasd
> (https://github.com/whjvenyl/fasd). Any idea how fasd could cause
> `;exit` to be interpreted as a valid command? Running `alias exit`
> shows that it's not aliased to anything.
>
> Weird that fasd causes `;exit` to be a valid command, and also that it
> causes bash to execute it when loading if it's the last thing in the
> history file. I'll raise a bug report on the fasd GitHub. Apologies
> for the erroneous report.
What you know is (more or less) that having ";exit" in the history file
causes "eval "$(fasd --init auto)"" to do an exit; you don't know that
that exit is due to exiting the specific characters of ";exit".
But as a start, you can just do
$ fasd --init auto
and see what that outputs, as it is likely that the exit you are seeing
is when "eval" executes the output of that command.
Dale
