3918795776 9926119
4118795776 9926119
23828480 9926120
223828480 9926120
423828480 9926120
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The first value should be the low part and the second value the high part.

---
print map { my ($m)=1<<hex($_)&11?' ':'';
$m.=substr('AHJPacehklnorstu',hex($_),1) }
split //,'2fde0abe76c36c914586c';
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I've imported your subroutine into my code... However, I'm not getting the what I think I should. If I've got it right, your routine should return the unix time in epoch seconds. However, doing a "gdate '+%s'" on my system shows the epoch seconds = 1045700265... a bit different to what the routine returns (see below). In my case, perhaps the unpack format of the low_long and high_long paramenters in the routine require a little alteration? BTW, reversing the order of $high and $low into &quad_to_epoch didn't help...

  print "$low $high ", &quad_to_epoch("$low $high"), "\n";
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low        high     &quad_to_epoch
2038398976 10599419 408804118300
2238398976 10599419 408804118300
2438398976 10599419 408804118300
2638398976 10599419 408804118300
2838398976 10599419 408804118300
3038398976 10599419 408804118300
3238398976 10599419 408804118300
3438398976 10599419 408804118300
3638398976 10599419 408804118300
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Regards, Stacy.

low_long is the least-significant 32 bits of the quadword date, while high_long is the most significant 32 bits. Make sure that you've unpacked things correctly--the code assumes that a VMS system put the bytes on disk, and in VMS endian-order. If that's not the case, then you'll have problems, of course.

Also, what are the dates supposed to be? Running the conversion on the first date gives me a date of Wed Dec 22 00:14:00 1993, which might well be correct. (Hard to say there)

Whilst parsing the VMS binary file, I have unpacked it with the template:

my $template="V2 v12";
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Where V/v are a long/short in "VAX" (little-endian) order. As I've already unpacked the data, there isn't a need to redo the unpack commands in the subroutine - just pass the low_long and high_long as is into the sub. There is one problem: the data in the binary file starts at 00:00:00 and ends at 23:40:40 for 1-1-2003. Printing out the output from the sub produces 11:00:00 through to 23:40:40 and _then_ 00:00:00 through to 10:59:40 for 2-1-2003! Regards, Stacy.