Tested on my 64 bit machine: 
use strict;
use warnings;

my $HIGH_BITS =  0xFFFFFFFF << 28;
printf "%X\n", $HIGH_BITS;
# prints: FFFFFFFF0000000
# each hex digit is 4 bits, 7 zeroes x4 = 28 bits left shifted

[download]
Usually displaying something like that in decimal would be nonsensical.

If you have an "all ones" in a 32 bit register, and you shift the contents 28 bits to the left, the high nibble will be "F". This will result in a negative value as viewed as a 2's complement number.

Update: jwkrahn came up with the ball on this one:
The trick is how to get Perl to display this value as signed integer.
On Windows the quoting is a bit different, But this works on my Windows machine.

>perl -le "my $HIGH_BITS = unpack "l", pack "l", 0xFFFFFFFF << 28; pri
+nt $HIGH_BITS; "
-268435456

[download]
A simple: print $HIGH_BITS on my machine yields: 1152921504338411520 which of course is not right. This looks like the unsigned decimal value.

In reply to Re: Difference between Perl and Java for << operator? by Marshall
in thread Difference between Perl and Java for << operator? by davidfilmer

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