Re: More info on sporadic OS/X crash

[Kim F. Storm]

"Piet van Oostrum" <address@hidden> writes:

> >>>>> "Eli Zaretskii" <address@hidden> (EZ) wrote:
> 
> >> From: "Piet van Oostrum" <address@hidden>
> >> Date: Wed, 28 Apr 2004 13:14:59 +0200
> >> 
> >> (gdb) frame 1
> >> #1  0x0012eeac in sys_select (n=126, rfds=0x38e7e4, wfds=0x0, efds=0x0, 
> >> timeout=0xbfffc770) at mac.c:2787
> >> 2787        return select(n, rfds, wfds, efds, timeout);    
> >> (gdb) print *timeout
> >> $3 = {
> >> tv_sec = 0, 
> >> tv_usec = 999996
> >> }
> >> 
> >> So this looks normal.
> 
> EZ> Well, yes and no: how come it's 999996 microseconds instead of a full
> EZ> second?  That is, why don't you see this instead?
> 
> EZ>  $3 = {
> EZ>    tv_sec = 1, 
> EZ>    tv_usec = 0
> EZ>  }
> 
> EZ> This higher frame in the backtrace:
> 
> >> #2  0x00119948 in wait_reading_process_input (time_limit=1, microsecs=0, 
> >> read_kbd=3506604, do_display=0) at process.c:4311
> 
> EZ> seems to imply that wait_reading_process_input was called to wait for
> EZ> 1 second and 0 microseconds, so where from did the small inaccuracy
> EZ> creep in?
> 
> There is some code just above the select that manipulates the usecs.
> (The ADAPTIVE_READ_BUFFERING stuff). But I think it always makes it a
> multiple of READ_OUTPUT_DELAY_INCREMENT, which this isn't. 
> 
> Then there's also timer_delay which gets calculated from something
> with the current time in it, so maybe that's it.

The Linux kernel adjusts the timeout value upon return from select to
contain the amount of time "remaining to the specified timeout".  This
is very useful in some situations.

Maybe OS/X does that too.

-- 
Kim F. Storm <address@hidden> http://www.cua.dk