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ntpq state

Von: Helmut Schneider (jumper99@gmx.de) [Profil]
Datum: 03.05.2009 19:40
Message-ID: <766392F1artrfU1@mid.individual.net>
Newsgroup: de.comp.os.unix.misc
Hi,

# ntpq -c "host localhost" -c rv
current host set to localhost
assID=0 statusf4 leap_none, sync_lf_clock, 15 events,
event_peer/strat_chg,
version="ntpd 4.2.4p5-a Fri Jan 16 01:56:19 UTC 2009 (1)",
processor="i386", system="FreeBSD/7.1-RELEASE-p5", leap, stratum=1,
precision=-19, rootdelay=0.000, rootdispersionG9.981, peer472,
refidÜFa, reftimeÍa85500.1a830613  Sun, May  3 2009 19:38:08.103,
poll=6, clockÍa8555e.0c33d84e  Sun, May  3 2009 19:39:42.047, state=4,
offset=-35.677, frequency=-45.456, jitter=1.789, noise=3.625,
stability=0.397, tai=0
#

Was bedeutet "state=4"? Ich denke "in sync", aber wo kann ich
Genaueres
nachlesen?

Danke und Gruß, Helmut

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