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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:
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
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.

"What do I want? I'm an American. I want more."

In reply to Epoch time in milliseconds: Is there a better way? by logan

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