How to remove bash key binding "x\000": self-insert

Question

bash -p shows

"x\000": self-insert

May be that this came in by a typo.

In order to repair, a normal bind command

bind "x":self-insert

or

bind '"x":self-insert'

has no effect. Therefore I try to remove the false binding by

bind -r "x\000"

It does effect nothing.

I have tried all variations with quotes and backslashes and $'\x00' and the like ('\0'), e.g.

bind -r "x"'\0'

without succsess.

But this was the solution --,for the first time:

bind -r ' "x" "$(echo -n -e "\x00")" '

For a second equal case this did not work.

The second case was because the normal command

bind '"x":self-insert'

caused again that bash -p shows

"x\000": self-insert

Is there really no unmistakenp way?

=====

== edit 08.12.2022 16:05

Thanks, but I need a way not to close the shell in order not to loose things.

. inputrc or other initiating files are not affected.

With my solution presented in the answer the unwanted binding is removed succsessfully.

The question now is how to make a new correct binding.

Even

bind "$(echo -n -e "\0170")":self-insert

is no wsy. All what I do results in

bash -p shows

"x\000": self-insert

I cannot ask you to help because you do not have such a corrupted shell.

( I do not have the courage to speculate if this is another craziness of bash )

Regards

Anton Wessel

Answer 1

The solution to is

bind -r "x" "$(echo -n -e "\x00")"

The problem why

 bind '"x":self-insert'

caused again that bash -p shows

"x\000": self-insert

is not yet solved. The reason is that I have not in this command to type "x" literally but in octal or decimal (in hex it cannot work because there "x" is involved).

I am trying how to insert "x" in decimal.