crond, hwclock, and clockspeed

[Lloyd Zusman]

I've got my linux box running with clockspeed, with
"/etc/leapsecs.dat", and with "/etc/localtime" pointing to a "right"
timezone file, namely "/usr/share/zoneinfo/right/America/New_York".

In most cases, my times are shown correctly.  For example, if I do
the test below, the times in the "Date:" header and in the body of
the message agree:

  (sntpclock 209.51.161.238 | clockview ; date) \
  | mail -s date ljz@...

However, it looks like "crond" is not taking the leapseconds into
account. The following is an excerpt from /var/log/cron:

  Sep 12 07:07:39 home crond[6187]: (ljz) CMD (/opt/bin/eachminute
    1>/dev/null 2>&1)
  Sep 12 07:08:39 home crond[6191]: (ljz) CMD (/opt/bin/eachminute
    1>/dev/null 2>&1)

I have restarted crond with an explicit "TZ=right/America/New_York"
in the environment, but cron still seems to be ignoring the
leapseconds.

Can anyone explain this and suggest a way to get "crond" to realize
the correct time?

Also, I have noticed the following behavior with the "hwclock" program:

  # (hwclock --set --date="$(date'+%m/%d/%y %H:%M:%S')"; \
      hwclock --show; date)
  Mon Sep 12 07:16:31 2005  -0.494631 seconds
  2005-09-12T07:16:52-0400
  # (export TZ=right/America/New_York; \
      hwclock --set --date="$(date'+%m/%d/%y %H:%M:%S')"; \
      hwclock --show; date)
  Mon Sep 12 07:17:50 2005  -0.494631 seconds
  2005-09-12T07:18:11-0400
  # (export TZ=America/New_York; \
      hwclock --set --date="$(date '+%m/%d/%y %H:%M:%S')"; \
      hwclock --show; date)
  Mon Sep 12 07:19:10 2005  -0.497663 seconds
  2005-09-12T07:19:09-0400

In other words, it seems like the "hwclock" program only shows the
correct time _without_ a "right" timezone, even though the rest of
my system seems to be operating correctly with "right" and with
"/etc/leapsecs.dat".

Can someone help me understand this?

Thanks in advance.



--
 Lloyd Zusman
 ljz@...
 God bless you.

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