Java bitwise operation

Question

I have this line of code

int b1 = 0xffff & (content[12]<<8 | 0xff & content[11]);

I have a bytearray (content[]) in little endian and need to recreate a 2 byte value. This code does the job just fine but prior to testing i had it written like this

int b1 = 0xffff & (content[12]<<8 | content[11]);

and the result was not right. My question is why is 0xff necessary in this scenario?

Because bytes are interpreted as signed integers, and this means, for example, that (byte) 0x80 gets (int) 0xffffff80. This is called sign extension. — Ingo
– Ingo, Commented Apr 4, 2013 at 18:26

Ted Hopp · Accepted Answer · 2013-04-04 18:26:11Z

12

The 0xff is necessary because of a confluence of two factors:

All integer types in Java are signed
All bitwise operators promote their arguments to int (or long, if necessary) before acting.

The result is that if the high-order bit of content[11] was set, it will be sign-extended to a negative int value. You need to then & this with 0xff to return it to a (positive) byte value. Otherwise when you | it with the result of content[12]<<8, the high-order byte will be all 1s.

answered Apr 4, 2013 at 18:26

Ted Hopp

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1 Comment

John Over a year ago

Thx for the reply! I would not have guessed content[11] would be casted to int with high-order bit set. I thought when i combine two values with OR they would be used AS ARE on the bit level. Cheers

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