Subject: bug#12587: 24.2; Delayed startup, unresponsive Emacs in MS Windows when netlogon services is running in a domain

From: Eli Zaretskii <[email-masked]>
Date: Mon, 08 Oct 2012 17:47:15 +0200

> Date: Mon, 08 Oct 2012 16:34:18 +0200
> From: Mohammed Imaduddin Humayun <[email-masked]>
> CC: Eli Zaretskii <[email-masked]>
> 
> ========================
> Before Welcome Screen
> =======================

This part is clear: we are waiting for file security-related system
calls:

> #21 0x76821e1b in LookupAccountSidW () from C:\Windows\syswow64\advapi32.dll
> #22 0x010572d7 in lookup_account_sid (lpSystemName=0x0, Sid=0xce1bc8,
>      Name=0x88f7cf "humayun", cbName=0x88f7c8,
>      DomainName=0x88f3c8 "GEOINFORMATIK", cbDomainName=0x88f3c4,
>      peUse=0x88f8d0) at w32.c:495
> #23 0x0105b343 in get_name_and_id (psd=0xce1b98,
>      fname=0x88f970 "c:\\emacs-24.2.50\\etc\\GNU", id=0x88fbf4,
>      nm=0x88fd14 "/emacs-24.2.50/etc/GNU", what=2) at w32.c:3437
> #24 0x0105b43c in get_file_owner_and_group (psd=0xce1b98,
>      fname=0x88f970 "c:\\emacs-24.2.50\\etc\\GNU", st=0x88fbe0) at 
> w32.c:3467
> #25 0x0105b91c in stat_worker (path=0x164a5c2 "\\emacs-24.2.50\\etc\\GNU",
>      buf=0x88fbe0, follow_symlinks=1) at w32.c:3689
> #26 0x0105bf49 in stat (path=0x39e9e50 "c:/emacs-24.2.50/etc/GNU",
>      buf=0x88fbe0) at w32.c:3865
> #27 0x01067a18 in Ffile_exists_p (filename=59080209) at fileio.c:2506
> #28 0x010d48f1 in init_callproc () at callproc.c:1582

As you see, these calls are originated from a function that needs to
establish whether a certain file exists.  In a Windows domain, these
calls go to the domain controller, and are evidently very slow when
the Netlogon service is up and running.  I think I can find a solution
for this.

> ===================
> After Welcome Screen
> ===================
> Attaching to program `C:\emacs-24.2.50\bin\emacs.exe', process 6444
> [New Thread 6444.0x2104]
> [New Thread 6444.0x2ab4]
> [New Thread 6444.0x1338]
> [New Thread 6444.0x2cf0]
> [New Thread 6444.0x2fe0]
> [New Thread 6444.0x1708]
> [New Thread 6444.0x2fbc]

I don't understand why we have 7 threads, while there are only 3 in
the previous case.  Can you please run Emacs under GDB to begin with,
and show where these threads are started.  Like this:

   cd \path\to\emacs.exe
   gdb ./emacs.exe
   (gdb) break CreateThread

(GDB will ask whether to make this breakpoint pending on future shared
library load; answer YES.)

   (gdb) commands
   backtrace
   continue
   end
   (gdb) run -Q

Now, whenever another thread is created, GDB will stop Emacs and print
the backtrace, then Emacs will continue.  Please show the results
here.

The main thread is just sitting in sys_select, waiting for input:

> #13 0x010551a1 in sys_select (nfds=1, rfds=0x88f6c0, wfds=0x0, efds=0x0,
>      timeout=0x88f6b0, ignored=0x0) at w32proc.c:1270
> #14 0x0104cd4e in wait_reading_process_output (time_limit=28, nsecs=0,
>      read_kbd=-1, do_display=1, wait_for_cell=56346650, wait_proc=0x0,
>      just_wait_proc=0) at process.c:4647
> #15 0x010fbc2a in sit_for (timeout=112, reading=true, do_display=1)
>      at dispnew.c:5977
> #16 0x010095c4 in read_char (commandflag=1, nmaps=2, maps=0x88f960,
>      prev_event=56346650, used_mouse_menu=0x88fa38, end_time=0x0)
>      at keyboard.c:2707
> #17 0x0101ccb2 in read_key_sequence (keybuf=0x88fbc0, bufsize=30,
>      prompt=56346650, dont_downcase_last=0, can_return_switch_frame=1,
>      fix_current_buffer=1) at keyboard.c:9312
> #18 0x01005f89 in command_loop_1 () at keyboard.c:1487

This means Emacs is idle and waits for the user to type something, or
for some Windows event to come in.  This is normal.

I see that I misinterpreted your original report about the startup
time line.  I now understand that after the welcome screen Emacs is
responsive, and the slowdown is before the welcome screen.

So please repeat what you did, but do it once before the GUI shows up,
and then again between the time the GUI shows up and the time Emacs
shows its welcome screen.  (If you know to which of these two time
instances belongs the first backtrace you show above, you need only to
produce the backtrace for the other time instance.)

Thanks.