bitwise AND in java with operator "&"

Question

I just read this code following:

byte[] bts = {8, 0, 0, 0};
if ((bts[i] & 0x01) == 0x01)

Does this do the same thing as

if (bts[i] == 0x01)

If not,what's the difference between them?

And what is the first way trying to do here?

score 13 · Accepted Answer · answered Jul 31 '13 at 10:56

13

No, it doesn't.

if(bts[i] == 0x01)

means if bts[i] is equal to 1.

if((bts[i] & 0x01) == 0x01)

means if the least significant bit of bts[i] is equal to 1.

Example.

bts[i] = 9 //1001 in binary

if(bts[i] == 0x01) //false

if((bts[i] & 0x01) == 0x01) //true

answered Jul 31 '13 at 10:56

Armen Tsirunyan

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@Johnny: E.g., **all odd numbers** match the `& 0x01` test, but only `0x01` matches the `== 0x01` test. – T.J. Crowder Jul 31 '13 at 10:59
It seems that all odd numbers will return true here. – Johnny Chen Jul 31 '13 at 11:04
1

@JohnnyChen: Do I hear an echo? ;-) – T.J. Crowder Jul 31 '13 at 11:07

score 2 · Answer 2 · answered Jul 31 '13 at 10:56

2

(0x1001 & 0x01) == 0x01, but

0x1001 != 0x01

answered Jul 31 '13 at 10:56

gefei

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score 2 · Answer 3 · answered Jul 31 '13 at 10:56

2

No, it doesn't, the first will check only the last bit - if it's 1, it will return true regardless of the others.

The second one will return true if only the last bit is 1.

answered Jul 31 '13 at 10:56

Luchian Grigore

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score 1 · Answer 4 · answered Jul 31 '13 at 11:00

No, it's not the same thing. 0x01 is just 1. Now,

if (bts[i] == 0x01)

checks if bts[i] is equal to 1.

if ((bts[i] & 0x01) == 0x01)

Checks if the last (least significant) bit of bts[i] is equal to 1. In the binary system, all odd numbers have the last bit equal to 1. So, if ((bts[i] & 0x01) == 0x01) is basically checking, if the number in bts[i] is odd. It could be written as if (bts[i] % 2 == 1), too.

bitwise AND in java with operator "&"

4 Answers4