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?
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4 Answers
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
3 Comments
T.J. Crowder
@Johnny: E.g., all odd numbers match the
& 0x01 test, but only 0x01 matches the == 0x01 test.
Johnny Chen
It seems that all odd numbers will return true here.
T.J. Crowder
@JohnnyChen: Do I hear an echo? ;-)
(0x1001 & 0x01) == 0x01, but
0x1001 != 0x01
Comments
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.
Comments
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.
answered Jul 31, 2013 at 11:00
Adam Stelmaszczyk
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