Integer to hex string "&" ?

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Hi..

I've seen a method like "Integer.toHexString(i & 0xff);" (i is an initially declared integer).
I wonder what part does the "&" and "0xff" take ?

Does this "&" act like the "+" or something else ?
And what's the computing that has been ran between the brackets ? I mean the behavior of (integer value & hexadecimal value)...

I think I've sufficiently described my problem..

Thanks in advance !

Posted 28-Apr-14 2:44am

Mark

Updated 28-Apr-14 2:49am

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CPallini · Accepted Answer · 2014-04-28T02:51:00

Solution 1

& is the bitwise AND operator.
0xff is the hexadecimal representation of binary 11111111 (decimal 255).
Hence:
(i & 0xff) extracts the least significative byte from integer number i.

Posted 28-Apr-14 2:51am

CPallini

Comments

Mark 28-Apr-14 9:01am

"extracts the least significative byte from integer number i."

Sorry. I think I dont get this part. Whats the meaning of "extracting least significative byte" ? Can I have a further description please ?

CPallini 28-Apr-14 9:19am

In order to keep it short, suppose your system has 16 bit integers, then 0xff is represented by the binary 
0000000011111111 
If you AND any integer with such a number, only the 8 rightmost bits survive. 
For instance, if i=1070, that is 0000010000101110, then 
0000010000101110 & 0000000011111111 = 0000000000101110. 
(you may easily verify it using AND truth table).

Mark 28-Apr-14 9:35am

Oh. So the (i & 0xff) converts both i and 0xff integers into a binary manner and match each bits with AND operation !

By the way, can I use other logical operations such as "or, xor, not, nand" with toHexString method ?
If it's so what characters am I suppose to use instead of ampersand ?

phil.o 28-Apr-14 9:57am

It does not convert; what you have to understand is that your computer stores and manipulates integers in their binary form, regardless of the way it presents them to you (under integer form 255, under hexadecimal form 0xFF, etc...).
So what this operation does is to compute the bitwise and operation between both numbers, and returns the result. There is not any conversion in the process, just a legacy bitwise AND operation on two integers.

phil.o 28-Apr-14 9:58am

AND = &
OR = |
XOR = ^
NOT = ~

CPallini 28-Apr-14 10:06am

Of course it doesn't 'convert': it internally stores every number in binary format (after all, it is a computer :-) ), decimal and hexadecimal representations are just used to simplify the human interaction. The compiler provides this facility.
You may find the other bitwise operator here:
http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html

Mark 28-Apr-14 10:38am

Thanks :)

Sergey Alexandrovich Kryukov 28-Apr-14 10:54am

No. It does not "convert" anything. All numeric objects are always "binary".
—SA

Mark 28-Apr-14 10:01am

OK Thanks.. And isnt there any character for "NAND" ?

CPallini 28-Apr-14 10:07am

Nope, however, you now, NAND(a,b) = NOT(AND(a,b)) :-)

Mark 28-Apr-14 10:39am

Thanks. :)

Sergey Alexandrovich Kryukov 28-Apr-14 10:56am

5ed.
OP is advised to some programming manual before asking questions.
—SA

OriginalGriff · Accepted Answer · 2014-04-28T02:51:00

'&' acts like '+' in that it's an operator - but it does a different function.
It's a Binary AND operator: so it works on individual bits of the input numbers to generate a result.
For each bit it works out what the result will be independently - it generates a "1" if both input bits are "1" and a "0" for all other combinations:

0xFF is slightly different - it's a constant value, but instead of being in Decimal (the base 10 that you normally use) it's an Hexadecimal - base 16.
In base 10 you have ten digits:

0 1 2 3 4 5 6 7 8 9

In Base 16 that ere (surprise!) sixteen:

0 1 2 3 4 5 6 7 8 9 A B C D E F

So "0xFF" or "0xff" (they are the same value) is the same as 255 in base 10, but a lot easier to visualise as a binary number - 11111111 in base 2.

So "i & 0xff" is just the bottom eight bits of i