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Subject: Why am I getting a popup window?
From: Paul Stodghill <stodghil AT cs DOT cornell DOT edu>
Message-ID: <yz1vgp2ugfm.fsf@milhouse.cs.cornell.edu>
Date: 22 Mar 2001 09:16:01 -0500
Lines: 45
User-Agent: Gnus/5.0808 (Gnus v5.8.8) XEmacs/21.2 (Urania)
MIME-Version: 1.0
Content-Type: text/plain; charset=us-ascii

Under cygwin and using cygwin gcc, I compile the following program,

        milhouse$ cat foo.c
        #include <stdio.h>
        #include <assert.h>
        main () { assert(0); }
        milhouse$ gcc -o foo.exe foo.c
        milhouse$ 

Now run it from within XEmacs compiled for cygwin,

        milhouse$ xemacs-21.2-b46 -V
        XEmacs 21.2  (beta46) "Urania" [Lucid] (i686-pc-cygwin) of Wed Mar 21 2001 on MILHOUSE
        milhouse$ xemacs-21.2-b46 -vanilla
        M-x shell-command
        Shell command: foo.exe

I get a popup error window that contains the message from the failed
assertion, "Failed assertion, 0, at line 3 of file foo.c".

According to spy++, foo.exe owns the window, not XEmacs.

How can I prevent the popup window from being creating, and just have
the message to sent to the XEmacs process?

Does this have something to do with the fact that subprocesses from XEmacs
are not full-blown ttys? (ptys?)

I do not get the popup window when I compile foo.exe with MinGW gcc or MSVS
cl.

I do not get the popup window when I run foo.exe from the "console" (ie,
Start -> Run -> bash.exe)

I do not see this behavior with XEmacs 21.1.13 (Andy's net release).

Using 1.1.8 or 20010316 versions of the cygwin1.dll doesn't change this
behavior.

CYGWIN=tty or CYGWIN!=tty doesn't change this behavior.

Compiling XEmacs with and without the debugging options doesn't change this
behavior.

Thanks.


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