localtime hour count

ady has asked for the wisdom of the Perl Monks concerning the following question:

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Re: localtime hour count by vkon (Curate) on Apr 08, 2006 at 14:09 UTC
I see following: `coli ~ # perl t.pl Sat Apr 8 14:09:05 2006 Sat Apr 8 14:09:10 2006 Elapsed: 0:0:5` [download] but on another machine: `D:\TESTS>perl 3 Sat Apr 8 18:10:21 2006 Sat Apr 8 18:10:26 2006 Elapsed: 3:0:5` [download] Obviously, this is just conversion of time onto your local timezone settings (mine is GMT+3, yours is GMT+1? CET?)	[reply] [d/l] [select]
Re^2: localtime hour count by ady (Deacon) on Apr 08, 2006 at 14:16 UTC
yep, I'm i GMT+1, so i'll buy your explanation here Tnx! -- allan update: `my ($h,$m,$s) = (gmtime($t2-$t1))[2,1,0];` [download] will DWIM -- allan	[reply] [d/l]
Re^3: localtime hour count by nothingmuch (Priest) on Apr 09, 2006 at 10:18 UTC
This is still confusing durations with dates. See my post below. If the epoch was at 00:03:35 instead of 00:00:00 (and it might be on some scary operating system that someone just happenned to port perl to ;-) then your time delta would be much harder to explain =) -nuffin zz zZ Z Z #!perl	[reply]
Re: localtime hour count by Herkum (Parson) on Apr 08, 2006 at 15:04 UTC
Testing out your code, this is what happens. You subtracting the number of seconds from epoc for $t1 from the number of seconds from epoc for $2. `$t1 = 1000 $t2 = 1005 $sec_diff = $t2 - $t1; print "Time elapsed $sec_diff\n";` [download] When you call localtime it returning a date time format for the seconds for epoc based upon the time for WHERE YOU ARE! `print "GMT time is " . gmtime($sec_diff); # Thu Jan 1 00:00:02 1970 print "Local time is " . localtime($sec_diff); # Thu Jan 1 19:00:02 1970 - for me at least.` [download] Note: I realized someone else had mentioned the gmtime but I just wanted to explain it further.	[reply] [d/l] [select]
Re: localtime hour count by rjsaulakh (Beadle) on Apr 08, 2006 at 14:31 UTC
<i have tried something very rudimentary hope u can refine and work on it let me know if you have something better `#!/usr/lib/perl -w use strict; use warnings; my $t1 = time(); print "$t1 \n"; sleep 5; my $t2 = time(); ; print ($t2) ; my $diff = $t2 - $t1; my $mDiff = int($diff / 60); my $sDiff = sprintf("%02d", $diff - 60 * $mDiff); print "diff = $diff -> $mDiff\:$sDiff\n";` [download]	[reply] [d/l]
Re^2: localtime hour count by ady (Deacon) on Apr 08, 2006 at 14:43 UTC
Thanks rjsaulakh See my update above for another fix of the issue. Best regards, allan	[reply]
Re: localtime hour count by nothingmuch (Priest) on Apr 09, 2006 at 10:09 UTC
The problem is that when you subtract `$t1` from `$t2` you wind up with a number, that is incidentially both the duration that passed, in seconds, *and* a valid epoch number representing a certain date, in this case 5 seconds after the "epoch", which is January 1st, 1970 GMT on UNIX systems. An epoch date is the number of seconds that passed since the epoch (essentially also a form of a duration, but with a well defined base value). localtime accepts an epoch number representing a date and then calculates that date, converts it to the local time zone, and provides the separate fields (something like the Jan 1, 1970, 1:00:05, that is 5 seconds after the epoch, converted to your timezone). Even if you use gmtime this would still not make sense, because you are talking about a duration, and not a date. Use DateTime to subtract the two dates, play with duration sanely, and all that, or use simple modular arithmetics: `my ( $h, $m, $s ) = do { use integer; ( ($tdelta / (60 * 60)) % 24, ($ +tdelta / 60) % 60, $tdelta % 60 ) };` [download] but note that this conversion is lossy. You can also use Time::Duration::Object, which is convenient for formatting durations for users as '2 hours, 5 minutes' or '2 minutes, 10 seconds', that is - rounded to a certain number of units. -nuffin zz zZ Z Z #!perl	[reply] [d/l] [select]


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