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# swift byte

Posted on 2004-08-30
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Hi,

I have this easy question...but keep bugging so long...I don't know how exactly ">>" or "<<" will work...
I have a byte value...
Assume I have a value 0xAA and I would like to get two 0xA...how can I do so??  How can I switch my variables???

Xenia
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Question by:xenia27
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Assisted Solution

girionis earned 1000 total points
ID: 11929164
It is "shift" not "swift" :)

Shifting left means multiplying by 2, shifting right means dividing by two. To left-shift a number simply do:

13 << 1

http://java.sun.com/docs/books/tutorial/java/nutsandbolts/bitwise.html
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Expert Comment

ID: 11929173
> Assume I have a value 0xAA and I would like to get two 0xA...how can I do so??  How can I switch my variables???

To multiply it by two just do:

int number =  0xAA;
System.out.println((number << 1));
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Accepted Solution

thomas908 earned 1000 total points
ID: 11929180
Left-shift << (JLS §15.19)
bits are shifted to the left based on the value of the right-operand
new right hand bits are zero filled
equivalent to left-operand times two to the power of the right-operand
For example, 16 << 5 = 16 * 25 = 512
Decimal 16            00000000000000000000000000010000

Left-shift 5     00000000000000000000000000010000
fill right     0000000000000000000000000001000000000

the sign-bit is shifted to the left as well, so it can be dropped off or a different sign can replace it

Right-shift >> (JLS §15.19)
bits are shifted to the right based on value of right-operand
new left hand bits are filled with the value of the left-operand high-order bit therefore the sign of the left-hand operator is always retained
for non-negative integers, a right-shift is equivalent to dividing the left-hand operator by two to the power of the right-hand operator
For example: 16 >> 2 = 16 / 22 = 4
Decimal 16       00000000000000000000000000010000

Right-shift 2      00000000000000000000000000010000
fill left      00000000000000000000000000000100
discard right  00000000000000000000000000000100  -> Decimal 4

Decimal -16      11111111111111111111111111110000

Right-shift 2      11111111111111111111111111110000
fill left      1111111111111111111111111111110000
discard right  11111111111111111111111111111100  -> Decimal -4

Unsigned right-shift >>> (JLS §15.19)
identical to the right-shift operator only the left-bits are zero filled
because the left-operand high-order bit is not retained, the sign value can change
if the left-hand operand is positive, the result is the same as a right-shift
if the left-hand operand is negative, the result is equivalent to the left-hand operand right-shifted by the number indicated by the right-hand operand plus two left-shifted by the inverted value of the right-hand operand
For example: -16 >>> 2 = (-16 >> 2 ) + ( 2 << ~2 ) = 1,073,741,820
Decimal 16       00000000000000000000000000010000

Right-shift 2      00000000000000000000000000010000
fill left      00000000000000000000000000000100
discard right  00000000000000000000000000000100  -> Decimal 4

Decimal -16      11111111111111111111111111110000

>>> 2              11111111111111111111111111110000
fill left      0011111111111111111111111111110000

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Expert Comment

ID: 11929182
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Author Comment

ID: 11929183
sorry~  keep thinking "switch" insteading of "shift"...^^
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Expert Comment

ID: 11929187
>>Assume I have a value 0xAA and I would like to get two 0xA...how can I do so

Sorry didn't get it.
Could u plz elaborate what u exactly mean by two oxA.

Do u mean u want to assign it to another variable?
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Expert Comment

ID: 11929190
For bytes you do exactly the same:

byte b = (byte) 0xAA;
System.out.println((b << 1));
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Expert Comment

ID: 11929193
thomas908 I think xenia means to multiply it by two.
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Expert Comment

ID: 11929203
>>thomas908 I think xenia means to multiply it by two.

Then ur first post would do the trick
;)
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Author Comment

ID: 11929210
OK...this is what I mean...
Here is my situation...I enter a string "AAAAAAAA" and it will be stored into 4 bytes....then, I try to get this data, and I receive "-86" as I print out.  And I know "-86" equals to "FFFFFFAA"...so how can I transfer this value to the right value I need??

Hope this is clearer~  @@
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Expert Comment

ID: 11929228
Which datatype are u using?
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Expert Comment

ID: 11929230
Ur bits may be getting truncated getting because u may be using a smaller data type.
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Author Comment

ID: 11929232
the one I use to receive the information is a byte value...
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Expert Comment

ID: 11929235
>>so how can I transfer this value to the right value I need??

I think rather than converting u'll have to save ur bits from getting truncated by using a large enough datatype
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Expert Comment

ID: 11929237
xenia, a byte is 8 bits long, meanign the the highest value you can store is 1111. The "AA" value you are trying to store is 10101010 which is more than 1111, so the value of the byte becomes negative.
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Expert Comment

ID: 11929238
byte can store only 8 bits.
HAve u tried with int?
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Expert Comment

ID: 11929240
Bah, sorry the highest values is 11111111 (I need my coffee soon) :(
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Expert Comment

ID: 11929242
oops! just a few seconds late ;)
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Author Comment

ID: 11929264
not sure why...i can the byte variable into "int" variable..and I still have "-86"...
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Expert Comment

ID: 11929267
Can you show us your code?
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Expert Comment

ID: 11929270
can u post ut code/
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Expert Comment

ID: 11929274
If you do this:

System.out.println(((int) 0xAA));

you will get 170.
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Expert Comment

ID: 11929284
A is 1010 in binary, AA is 10101010 which is 128 + 32 + 8 + 2 = 170
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Author Comment

ID: 11929285
OK...I just realize why I still have "-86"...since the buffer I try to store my information into is a byte array...so when I parse the data, I still have "-86"...so is there any way I can get the correct value???? @@
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Expert Comment

ID: 11929293
Can u make it an int array?
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Expert Comment

ID: 11929296
U have to save the bits from getting truncated. Once they get truncated u can't get them back
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Author Comment

ID: 11929297
sorry...cannot do this...@@
it's a huge program...if I do so, I need to make tons of changes...
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Expert Comment

ID: 11929298
The byte array stores each character separately, not two together. The String "AAAAAAAA" will result in a byte array of length 8. Each location will have the character A which is represented as 0000000000001010. How do you store the data? Can you post some sample code?
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Expert Comment

ID: 11929299
So the idea is use a large enough datatype.
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Expert Comment

ID: 11929304
>>The byte array stores each character separately, not two together. The String "AAAAAAAA" will result in a byte array of length 8. Each location will have the character A which is represented as 0000000000001010. How do you store the data? Can you post some sample code

girionis is right. i missed the point
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Expert Comment

ID: 11929315
xenia please post the code where you do the byte stuff.
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Author Comment

ID: 11929318
OK...here is what I do...
First, I try to send my commands to a machine...so I send my string..."AAAAAAAA"
then, I need to convert the string to bytes...so I do this "tmpStr.getBytes()"...then, the machine returns a succeed signal...
so I try to update my program...then, what my program do is receive these data with a byte array...a buffer, then, I parse the information...so for this case, I need 4 bytes to store this string...so I have another byte array...byteArray[4]...and I would like to get the original string from this byte array...

Any idea how?
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Expert Comment

ID: 11929332
> .so for this case, I need 4 bytes to store this string.

No you do not. You need 8 bytes to store the string since you have 8 characters in a string.
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Expert Comment

ID: 11929334
If you do getBytes() on the string this will result in a byte array with length *eight not four*.
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Expert Comment

ID: 11929336
ur string is 8 bytes long
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Expert Comment

ID: 11929340
Xenia, do not confuse the *byte* value 0xAA with the *string* value "AA". They are *not* the same.
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Author Comment

ID: 11929343
OK...I check my program....I'm wrong...I have an array with length 8... Sorry~
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Expert Comment

ID: 11929359
I think I know what went wrong in your mind. You have confused the unicode characters of java with the string characters. The "A" character is represented as unicode (2 bytes - 16 bits) but the "A" character is "one" character not two. So the string "AA" is "two" characters of 16 bits each (32 bits in total) and not one character.
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Expert Comment

ID: 11931958
>>
OK...I just realize why I still have "-86"...since the buffer I try to store my information into is a byte array...so when I parse the data, I still have "-86"...so is there any way I can get the correct value???? @@
>>

If the buffer contains 0xAA, you can mask it, e.g.

byte[] buffer = { (byte)0xAA };
int value = buffer[0] & 0xFF;
System.out.println(value); // prints 170
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Author Comment

ID: 11938253
OK...what I try to do is try to print out "AA", which is a hex number, as a string...not int (170) or byte value...I totally understand what I have is hex value and "A" character in "AA" is part of the whole value...so that is why I wanna know how can I get two part of whole value separately by shifting byte...This is what I mean...sorry for misunderstanding...but since the original value in my buffer is incorrect...I don't know how can I do so...^^
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Expert Comment

ID: 11939257
Try this:

int number = 0xAA;
System.out.println(Integer.toHexString(number));
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Expert Comment

ID: 11939638
>>but since the original value in my buffer is incorrect...I don't know how can I do so..

What do you mean by that exactly?
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