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Time in Milliseconds

Posted on 1999-01-27
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Last Modified: 2008-03-17
I'm using a perl script to time things with code like $startmain=time; at the beginning and at the end of what I want to time.  Time seems to be limited to a minimum value of seconds, while I would like to measure down to the millisecond.  Is there a way to do this?

One oddment I found also is that when I want to establish the date and I use gmtime(time)=($var ...), I have to increment the month by 1 to get the current month.  Does anyone know why this is?
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Question by:wilfrieds
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by:ahoffmann
ID: 1210327
use perl's times builtin command (see Camel book).

> I have to increment the month by 1 to get the current month
'cause the month is returned in C-style indexing: 0..11
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by:wilfrieds
ID: 1210328
ahoffmann : times seems to provide me only with CPU times, where for this particular reason I want the total lapse time of the script - is there something else you are aware of?
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by:ozo
ID: 1210329
perldoc -q 'time.*second'

=head1 Found in perlfaq8.pod

=head2 How can I measure time under a second?

In general, you may not be able to.  The Time::HiRes module (available
from CPAN) provides this functionality for some systems.

In general, you may not be able to.  But if your system supports both the
syscall() function in Perl as well as a system call like gettimeofday(2),
then you may be able to do something like this:

    require 'sys/syscall.ph';

    $TIMEVAL_T = "LL";

    $done = $start = pack($TIMEVAL_T, ());

    syscall( &SYS_gettimeofday, $start, 0)) != -1
               or die "gettimeofday: $!";

       ##########################
       # DO YOUR OPERATION HERE #
       ##########################

    syscall( &SYS_gettimeofday, $done, 0) != -1
           or die "gettimeofday: $!";

    @start = unpack($TIMEVAL_T, $start);
    @done  = unpack($TIMEVAL_T, $done);

    # fix microseconds
    for ($done[1], $start[1]) { $_ /= 1_000_000 }

    $delta_time = sprintf "%.4f", ($done[0]  + $done[1]  )
                                            -
                                 ($start[0] + $start[1] );

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by:wilfrieds
ID: 1210330
thanks ozo
i 'll accept your answer if you care to post your comment as an answer
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Accepted Solution

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ozo earned 50 total points
ID: 1210331
If you're trying to benchmark a piece of code, you might also
use Benchmark;
which lets you run the code a number of times to get a more precise timing.
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