Bitwise XOR java long

Question

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?

Accepted Answer

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.

Bitwise XOR java long

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