java bitwise operators and the equal character; compound operators

Question

I'm a bit confused:

long v = 0;
v <<= 8; 
v |= 230;

I know << is the signed left shift operator and | Bitwise the inclusive OR but I'm confused to what the equals does?

So fist v is 0. So << doesn't have any effect? Then it equals 1000 but what happens then?

edit: I've edited the title so others might better find this question: added "compound operators"

It has an effect. `v = v << 8` and then `v = v | 230`. – Maroun Jul 25 '13 at 12:56 — Maroun, Jul 25 '13 at 12:56

score 5 · Answer 1 · answered Jul 25 '13 at 12:55

They're compound operators, like += and -= are. They do the operation, and then assign the result back to v.

Basically:

v <<= 8;

is in effect

v = v << 8;

And similarly

v |= 230;

is in effect

v = v | 230;

You can see the parallel with += and -=:

v += 1;

is effectively

v = v + 1;

score 5 · Accepted Answer · answered Jul 25 '13 at 12:55

5

There are somewhat like +=.

For example x+=3 means add 3 to x; store to x.

v <<= 8;

left-shifts v 8 bits, and stores to v, functionally equivalent to v=v << 8.

v |= 230;

does a bitwise OR with 230 and stores back to v, equivalent to v=v | 230.

Now, due to performance constraints and optimizations this operation may be done in place at a low level.

answered Jul 25 '13 at 12:55

nanofarad

@Thomas No problem. Please accept the answer that helped you most using the checkmark next to it. – nanofarad Jul 25 '13 at 13:04
2

I have to wait a few minutes before I can; but I will :) – Thomas Jul 25 '13 at 13:07
After the execution `v=v | 230` what will be the `v` value ?? – Suresh Atta Jul 25 '13 at 13:09
@sureshatta Just try it and see. It depends on the original `v` value – nanofarad Jul 25 '13 at 13:43

score 0 · Answer 3 · answered Jul 25 '13 at 12:56

0

Basically, this:

v <<= 8; 
v |= 230;

is equivalent to this:

v = v << 8; 
v = v | 230;

answered Jul 25 '13 at 12:56

jh314

3 Answers3