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

Hello monks,

I may have found a bug but don't know how to tell...

With perl 5.8.4, I run this code and get: 
use Time::Local;

my $hour= 2;
my $test = timelocal(0,0,$hour,1,3,2012);
my @test2 = localtime($test);
print "hour $hour vs $test2[2]\n";

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hour 2 vs 2

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If I run this identical code under perl 5.8.7, it outputs: 
use Time::Local;

my $hour= 2;
my $test = timelocal(0,0,$hour,1,3,2012);
my @test2 = localtime($test);
print "hour $hour vs $test2[2]\n";

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hour 2 vs 3

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If I run this 2011 modified code under 5.8.7, it outputs: 
use Time::Local;

my $hour= 2;
my $test = timelocal(0,0,$hour,1,3,2011);
my @test2 = localtime($test);
print "hour $hour vs $test2[2]\n";

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hour 2 vs 2

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Am I going crazy? What is going on?!? AHhhh!

-Alan

Replies are listed 'Best First'.
Re: Timelocal, Localtime Bug?
by JavaFan (Canon) on Mar 29, 2012 at 23:32 UTC
    5.8.7 isn't the latest release in the 5.8 series, there have been 2 more. Beside that, 5.8.x is no longer supported. 5.10.x is no longer supported either. I don't expect a new release in the 5.12 series to happen either.

    I suggest checking what your code does in 5.14, or in 5.15.9. If that prints what you think it should print, consider the bug to be fixed.

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      The problem is that I am constrained to the installed version, and it would be a heck of a fight to get the administrators to update to a newer version.
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        So, if there's a bug, and it got fixed, it would be utterly useless to you anyway?
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Re: Timelocal, Localtime Bug?
by jwkrahn (Abbot) on Mar 29, 2012 at 21:53 UTC 
    $ perl -le'my $test = localtime( 0, 0, $month, 1, 31, 2012 ); print $t
+est'
Too many arguments for localtime at -e line 1, near "2012 )"
Execution of -e aborted due to compilation errors.

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    Your code won't even compile, let alone run.

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[d/l]

      It looks like it might run if you flip localtime and timelocal with each other (I'm guessing timelocal takes the arguments localtime is currently taking and vice-versa).

      That said, I have no idea concerning the original question. ><

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[d/l]
[select]
        Sorry guys! My development machine is not connected to the internet, so I had to type the chunk of applicable code and I mixed up timelocal and localtime. I fixed the code samples in the original post, so now my original question actually applies (also put 3 instead of 31... I hate fat fingers today). Sorry!
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Re: Timelocal, Localtime Bug?
by Riales (Hermit) on Mar 29, 2012 at 22:16 UTC

    This is from the documentation on Time::Local:

    $time = timelocal($sec,$min,$hour,$mday,$mon,$year);

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    It looks like you're putting $month where $hour is supposed to go?

    EDIT: Nevermind, it looks like you're consistently checking the hours anyways. The 3 instead of 2 makes me think that it has something to do with DST (2 AM doesn't exist because we spring forward to 3 AM in the spring). Still, it doesn't happen on March 1st, so this doesn't seem quite correct...

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[d/l]
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      Sorry, I fixed the code chunk. I took it from my other machine and had fat fingers (mixing up localtime and timelocal...), so now with the code that's posted, the bug appears.
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Re: Timelocal, Localtime Bug?
by thundergnat (Deacon) on Mar 30, 2012 at 14:29 UTC

    I would suspect that the version of Time::Local installed on your machine is older than version 1.13.

    From the changelog:

    1.13    2006-08-09

...

- Fixed a bug that occurred around the DST change for Europe/London
  (and probably other time zones with a positive UTC offset). If
  given, the hour immediately after the change (2:00 AM for
  Europe/London), then the returned epoch was 3600 seconds too
  large. Reported by Roger Picton. See RT #11662.
[reply]

      I dare say you've exactly found the bug I'm experiencing. Thanks for the help!

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