Since Java integers are always signed and with sizes:

it seems to me the OP's Java code:

long HIGH_BITS = 0xFFFFFFFF << 28;
is ill-advised because (as has been pointed out multiple times in this thread) 0xFFFFFFFF << 28 is intrinsically a 32-bit value, while a Java long is 64-bits. Though the OP's code presumably "works", surely it ought to have been written as:
int HIGH_BITS = 0xFFFFFFFF << 28;

In case the OP is still listening, I won't be able to sleep at night until I learn the backstory as to why the Java code chose a long rather than an int. :)

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

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