logan has asked for the wisdom of the Perl Monks concerning the following question:
I'm working on a project that demands I pass the time since the epoch in milliseconds. Getting the value was simply a matter of using Time::HiRes::gettimeofday. Now I've got the seconds since the epoch plus the extra microseconds. My understanding was that the function would return the milliseconds, but the value returned has six digits, not three.
That quirk notwithstanding, my main issue is that my method of joining the two values together into a single number seems inelegant and I wonder if I'm missing something.
Here's my code:
It seems to me that I'm using three lines of code to do something that should be fairly basic and that using sprintf to trim microseconds into milliseconds is a complete hack. Moreover, the docs for gettimeofday() continually refer to the value returned as "milliseconds", which implies a three-digit number (thousandths of a second). I suspect that there's a better way to do this but I'm kind of stumped.use Time::HiRes qw(gettimeofday); $one = gettimeofday; print STDOUT "one = $one\n"; ($sec,$milli) = gettimeofday; print STDOUT "sec = $sec, milli = $milli\n"; $milli2 = sprintf("%.*s", 3, $milli); print STDOUT "sec = $sec, milli = $milli, milli2 = $milli2\n"; $time = join ('',$sec,$milli2); print STDOUT "time = $time\n"; Output: one = 1227582343.55258 sec = 1227582343, milli = 552702 sec = 1227582343, milli = 552702, milli2 = 552 time = 1227582343552
-Logan
"What do I want? I'm an American. I want more."
|
---|
Replies are listed 'Best First'. | |
---|---|
Re: Epoch time in milliseconds: Is there a better way?
by GrandFather (Saint) on Nov 25, 2008 at 03:37 UTC | |
by logan (Curate) on Nov 25, 2008 at 06:08 UTC | |
by Xilman (Hermit) on Nov 25, 2008 at 12:31 UTC | |
by logan (Curate) on Nov 25, 2008 at 20:34 UTC | |
by GrandFather (Saint) on Nov 25, 2008 at 06:42 UTC |
Back to
Seekers of Perl Wisdom