dagronlund dagronlund - 1 year ago 133
Java Question

What does value & 0xff do in Java?

I have the following java code:

byte value = 0xfe; //Signed for -2 and unsigned for 254
int result = value & 0xff;

Result is 254 when printed, but I have no idea how this code works. If the
operator is simply bitwise, then why does it not result in a byte and instead an integer? Any help would be appreciated.

Answer Source

It sets result to the (unsigned) value resulting from putting the 8 bits of value in the lowest 8 bits of result.

The reason something like this is necessary is that byte is a signed type in Java. If you just wrote:

int result = value;

then result would end up with the value ff ff ff fe instead of 00 00 00 fe. A further subtlety is that the & is defined to operate only on int values1, so what happens is:

  1. value is promoted to an int (ff ff ff fe).
  2. 0xff is an int literal (00 00 00 ff).
  3. The & is applied to yield the desired value for result.

(The point is that conversion to int happens before the & operator is applied.)

1Well, not quite. The & operator works on long values as well, if either operand is a long. But not on byte. See the Java Language Specification, sections 15.22.1 and 5.6.2.

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