999000000

[Howard Lowndes]

But wouldn't that have made it 36000 seconds even later, or am I not
thinking straight.

-- 
Howard.
LANNet Computing Associates - Your Linux people
Contact detail at http://www.lannetlinux.com

On Wed, 29 Aug 2001, Ian McCulloch wrote:

>
> On Wed, 29 Aug 2001, Martijn van Oosterhout wrote:
>
> > On Wed, Aug 29, 2001 at 08:03:33AM +1000, Howard Lowndes wrote:
> > > Slightly wrong
> > >
> > > $ date -u --date="1 jan 1970 + 1000000000 secs"
> > > Sat Sep  8 15:46:40 UTC 2001
> > > $ date  --date="1 jan 1970 + 1000000000 secs"
> > > Sun Sep  9 01:46:40 EST 2001
>
> This is 1 jan 1970 local time, which is 36000 seconds infront of UTC.
> try
> $ date -u --date="1 jan 1970 UTC + 1000000000 secs"
> Sun Sep  9 01:46:40 UTC 2001
> $ date --date="1 jan 1970 UTC + 1000000000 secs"
> Sun Sep  9 11:46:40 EST 2001
>
> >
> > Something is wrong here, look:
> >
> > $ date --date='Sep  8 15:46:40 UTC 2001' +%s
> > 999964000
> > $ date --date='Sep  9 01:46:40 2001' +%s
> > 999964000
>
> yep
>
> >
> > I worked it out like follows:
> >
> > $ perl -e 'print scalar(localtime(1_000_000_000)),"\n"'
> > Sun Sep  9 11:46:40 2001
> > $ date --date='Sep  9 11:46:40 2001' +%s
> > 1000000000
> > $ perl -e 'print scalar(gmtime(1_000_000_000)),"\n"'
> > Sun Sep  9 01:46:40 2001
> > $ date --date='Sep  9 01:46:40 2001 UTC' +%s
> > 1000000000
> >
> > So who is right?
> >
>
> I guess the 'real' zero is in UTC ?
>
> Cheers,
> Ian McCulloch
>
>