Zsh Mailing List Archive
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Re: Overriding "builtin"

Following up on this a bit, it seems that if your Zsh code is executing in a malicious environment (e.g. has done "function /usr/bin/sudo() { echo lol }") is to use a non-qualified path with the "=" prefix.

$ function /usr/bin/sudo { echo lol }

$ /usr/bin/sudo whoami

$ =/usr/bin/sudo whoami

$ =sudo whoami

Why does "=sudo" do the correct thing (assuming a sane $PATH, and executes /usr/bin/sudo), but "=/usr/bin/sudo" does the wrong thing (i.e., execute the function)?

Assume "builtin", "command", "exec", etc. have all been overwritten with functions.

Since the environment is malicious, $PATH also cannot be trusted -- I thought "=" might be a way to guarantee that an executable at a specific absolute path does get executed instead of something else (alias, function, autoloadable, etc) but it doesn't work when specifying the full path.

Zach Riggle

On Fri, Aug 12, 2022 at 3:54 PM Bart Schaefer <schaefer@xxxxxxxxxxxxxxxx> wrote:
On Fri, Aug 12, 2022 at 12:33 PM Zach Riggle <zachriggle@xxxxxxxxx> wrote:
> It would be nice if we could add a feature such that the "builtin" identifier cannot be overloaded.

This isn't really feasible, given that we have e.g. "disable builtin"
and "alias builtin=...".

That does point out that another approach to bypassing the function is
 disable -f builtin
which is pretty nice in that it leaves the function defined but
inaccessible.  Of course one can still "disable disable" as well.

I can't imagine why anyone would do this, but of course

disable -rm \*
disable -m \*

leaves the shell able to only to execute pipelines built from external
commands.  Preceded with a few "alias -g" of separators, you end up
limited to simple external commands.

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