This is because time is using a 32 bit integer and you are getting an overflow truncation like you do in say C. Here is a demo of what happens:

$time = time();
$odd  = time * 1000;
print "$time|$odd\n";

print "2**31 is ", 2**31, "\n";
$time1000 = $time *1000;
$time1000 %= 2**31;
print "Here is our number: $time1000\n";
print "End of *nix epoch is at: ", scalar gmtime (2**31-1)
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The maximum integer value that time will hold is 2**31 or 2147483648. As $time1000 is much bigger than this we get overflow tructation. We simulate this with modulus % and lo and behold produce the same number.

cheers

tachyon

s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print

So, I understand this is true. I'm O.K. with overflows. I can handle overflows. But why does

$var = time * 1000;
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overflow and

 
$var = time; $var *= 1000;
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not overflow?

I believe that on my system (Win2k, perl 5.6.1) integers are 32 bit, just like returned by time() I verified this with print ($var&(~0)) which gives me (2**32)-1.

Does perl automatically decide (not) to use BigInts? Any ideas?

Ira,

"So... What do all these little arrows mean?"
~unknown

time() is a Perl function, implemented in C with 32 bit (long int) math relevant to the unix epoch.

A perl scalar is an interesting beast (also implemented in C) that can hold a string or a number or a reference so is not directly comparable to a short, int, long, float, double, char in C. It is in fact a struct. Have a look at this code and perlman:perlguts. On the Win95 box in front of me Perl will represent up to 2**49 (562,949,953,421,312) as an integer before resorting to scientific notation.

for ( 0..64 ) {
    $var = 2**$_;
    print "Perl can do 2**$_ : $var\n";
}
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cheers

tachyon

s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print

Maybe the perl hackers (as apposed to the Perl hackers) can correct me: your typical integer (x86 arch) is going to be 32bits. time returns an integer, and 1000 is an integer, so perl's going to return the results as an integer too. But the largest number a signed integer can express is roughly 2 billion. Time just spilled over into the 1 billions fairly recently. So the result is overflowing, giving you the "inaccurate" answer.

See how Math::BigInt works out for you.

-Ducky

That's right. You can use a special build option (-Duse64bit, IIRC) on some platforms(including Linux) to get 64-bit ints, which I think go into the trillions without overflowing.

=cut
--Brent Dax
Perl 6 Configure pumpking
There is no sig.

 perl -e 'print(time, "\n"); print(time*1e3, "\n");'
1001715463
1001715463000
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