Bitwise XOR java long

问题

I am using Oracle Java 7.51 on Ubuntu 12.04, and trying to do this

long a = 0x0000000080000001 ^ 0x4065DE839A6F89EEL;
System.out.println("result "+ Long.toHexString(a));

Output: result bf9a217c1a6f89ef

But I was expecting result to be 4065de831a6f89ef, since ^ operator is a bitwise XOR in Java. Which part of Java specification am I reading wrong?

回答1:

You need an L at the end of the first integer literal:

long a = 0x0000000080000001L ^ 0x4065DE839A6F89EEL;

Otherwise it is an int literal, not a long (the leading zeroes being ignored). The ^ operator then promotes the first operand value from 0x80000001 to a long, but since the sign bit is set, the result of the promotion is 0xFFFFFFFF80000001L.

来源：https://stackoverflow.com/questions/22651465/bitwise-xor-java-long