Re: misbehavior in shell window with ksh

[Eli Zaretskii <eliz@gnu.org>]

> From: Stephen Berman <stephen.berman@gmx.net>
> Cc: mkupfer@alum.berkeley.edu,  emacs-devel@gnu.org
> Date: Mon, 01 May 2017 17:52:31 +0200
> 
> > Can you show the backtrace for the invocation of comint-output-filter?
> 
> After `M-o' in the recipe I did `M-: (debug-on-entry
> 'comint-output-filter) RET' (when I tried `M-x debug-on-entry RET' the string
> "> " was then inserted into the *shell* buffer).  This showed this backtrace:
> 
>    Debugger entered--entering a function:
>    * comint-output-filter(#<process shell> "> ")
> 
> I typed `q' then proceeded with `M-0' from the recipe, which produced
> the same backtrace.  Why isn't the caller of comint-output-filter shown?

What if you step into comint-output-filter with Edebug (as you already
seem to have a way of doing that), then type 'd' to produce a
backtrace?  Does that show who called comint-output-filter?

It could be that it is called by the process-filter mechanism, which
is in C.  But what we want to know is where does the 2nd arg of
comint-output-filter comes from, and why.

Thanks.