Signed right shift operator

Hello,
why -1 >> 1 get -1, and -2 >> 1 also get -1?
How does >> operator works on negative number?
thanks.

it does set 1's to the input
example if u have
1111 1110 which is -2 the operator
do the following
he set a new 1 at the first position and move the rest one step to the right.
1 111 1111 and the 0 goes out better example
1011 1111 >> 1 =
110111111

Similar Messages

Is signed right shift operator shows starnge working?

can anyone explain abt this stange behaviour !!!
int i = 1;
pBinInt("i >>= 10 ",i);
static void pBinInt(String s, int i) {
     System.out.println(s + ", int: " + i + ", bin: ");
     System.out.print("\t");
     for(int j = 31; j >=0; j--)
          if(((1 << j) & i) != 0)
               System.out.print("1");
          else
               System.out.print("0");
     System.out.println();
The signed right shift >> uses sign extension:if the value is
positive, zeroes are inserted at the higher-order bits; if the value is
negative, ones are inserted at the higher-order bits.
unsigned right shift >>>, which uses zero extension: regardless of the
sign, zeroes are inserted at the higher-order bits.
but output shows :
i >>= 10 , int: 1, bin:
00000000000000000000000000000001
shouldn't it be 00000000000000000000000000000000

You print the binary representation of i, which is 1, instead of i >> 10, which is 0.

Logical right shift ( ) operator

Hi,
I suspect a bug in the logical right shift operator >>> (as opposed to the arithmetic right shift >>).
The documentation says "0-s will be shifted from the left". It is not the case.
byte b = -128; // this means 0x80 in hex notation
byte b1 = (byte)(b >>> 4);I would expect b1 = 0x8 (0-s shifted in from the left) but the result is
b1 = 0xF8 instead, as if I had used the >> operator. The sign bit was shifted in from the left instead of 0-s.
Of course, there is a workaround:
byte b1 = (byte)((b >>> 4) & 0x0F); which is the same operation as using >>
If this is the case, what is the purpose of using >>> instead of >> ?
Anyway, the ONLY thing that frustrates me in the Java language that it does not support unsigned integers. It would be important for embedded systems where unsigned numbers are used more frequently.
Any suggestions?
Thank you,
Frank

sabre150:
It depends on the size of processor you use. The Java program runs on a 32- or 64-bit processor, so there is no problem promoting from byte to short or int. But I am talking to an 8-bit processor over a network. So I get a byte stream that has to be converted to numbers here on the Java side. The data types are various: byte or short, even sometimes two 4-bit nibbles (here comes the >>> operator), signed or unsigned. If I get a two-byte sequence, I have to convert it to a short, signed or unsigned. That means in Java promoting the byte to int, shift left the high byte, OR the low byte then cast it back to short if signed, or mask with 0xFFFF if unsigned. The resulting int can be converted to string in the usual way. With distinguished signed and unsigned types, this would be done by the compiler. But I come from the Assembler (and then Pascal/Delphi) world, so it is not a big problem for me.
I agree that this feature is used very rarely, I already wrote the conversion routines, so why bother? As a newcomer to Java (after struggling with C for years, I love it), I just wanted to see whether other people have a better and more efficient idea.
Thank you all for your replies!
Frank

JAVA (SHIFT OPERATOR)

Hi,
I really want to know SHIFT operator in Java like >> , << , >>>
Could anybody kindly help to explain??
And
byte a =-1;
a = (byte)(a>>>2);
why the output become -1??
thanks.

umm i did get them mixed up. preserved sign bit right shift is >>, pulling-zeros is >>>. so keep that in mind. (ie, 18>>2 == 10010>>2 == 11100 which is -4)
To calculate the resulting number from >>> right shifts, check this: http://www.janeg.ca/scjp/oper/shift.html
In response to your original question, it seems it's far more complicated than I at first thought. The -1 that you're getting is an odd (and rather interesting, imo) byproduct of modular arithmetic. what you wrote is exactly similar to this code (ie, my code mimics the castings that happen in your code):
        byte a = -1;
        int b = (int) a;
        int c = b>>>2;
        byte d = (byte) c;
        System.out.print(d);So what happens is A, which is a string of ones (ie 11111111) is casted to an int, B. In general, casting a byte to an int returns an int with exactly the same bits at the end with the sign bit repeated 24 times at the start (ie, 01101011-->(int)0000...00001101011). So B is 11111111111111...1111 (32 ones). Then we right shift B by 2 (we'd get the same answer right shifting it by 0, 1, 2, etc, all the way up to (and including) 24). Then we recast the int C to a byte. This is where the interesting thing happens: Depending how much we right shifted B to get C, C will be a certain number of zeros (call this Z) followed by (32-Z) ones. Apparently, in (mod 2^8) in two's complement all of these numbers are congruent to -1, so the cast gives negative one.
(remember that the int 127 cast to byte is 127, while the int 128 cast to byte is -128. conversions to a lower-bit type always happen mod 2^N, where N is the number of bits in the type we're casting to.)
BTW, thanks for this problem. i had a fun half hour figuring it out. hopefully it made some sense to you...
(EDIT: Flounder is right--you can just view the cast from int to byte as keeping the last 8 bits. much easier to think of it that way, though my definition does give some intuitions about WHY all this works.)
Edited by: a_weasel on Oct 13, 2008 8:00 PM

Right shift and c not working, Left shift and c works fine for a capital c

I have a Lenovo G560 laptop. Had this problem once before and computer tech fixed problem, but moved cities. Worked fine but recently had a windows update. Now when I press the right shift key and c nothing happens. Left shift and c displays a capital c. How do I restore the right shift c ?

hi richiestac,
You can try to remap the Right Shift Key so that it will function as Left Shift using Sharpkeys.
To do this:
1. Download and install Sharpkeys
2. Open Sharpkeys, click on Add and on the Map this key (From Key) , select the Right Shift key, then on the To this key (To key), select the Left Shift key.
- Link to picture
3. When finished, click Write to Registry and reboot.
As for the display driver installation, you need to uninstall first any signs of Intel HD Graphics from the Control Panel (Icon View) > Programs and Features. After that, open Device Manager > Display Adapters and uninstall the Intel Graphics (ensure you put a check on Delete the driver software for this device and try to re-install the Intel Onboard Graphics Driver.
Alternatively, you can skip all the procedures above and just press the OneKey Recovery button to restore the system to factory settings (note that all data will be wiped on the OS partition).
Regards
Did someone help you today? Press the star on the left to thank them with a Kudo!
If you find a post helpful and it answers your question, please mark it as an "Accepted Solution"! This will help the rest of the Community with similar issues identify the verified solution and benefit from it.
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Left-shift operator used on byte values

Hello,
I'm reviewing some problems, and I need some help. I have a program that has some code like the following:
byte y = 10; // 00001010 in binary
byte result = (byte) (y << 1);
System.out.println("result: " + result); // 20. Ok.
result = (byte) (y << 7);
System.out.println("result: " + result); // 0. Ok.
result = (byte) (y << 8);
System.out.println("result: " + result); // 0. Why???
// I was expecting a shift of 0 bits because the
// right-hand operand is equal to the number of
// bits for the size of the result type--in this case
// 8 bits for a byte.
// 8 % 8 = 0 number of bits for the shift.
result = (byte) (y << 6);
System.out.println("result: " + result); // -128. Ok.
result = (byte) (y << 10);
System.out.println("result: " + result); // 0. Why???
// Shouldn't it be 2 bits for the shift?
// That is, 10 % 8 = 2.
// I was expecting 40 as the the answer for this one.I understand that for binary operations that the operands will be promoted to at least int types before execution occurs, but I still don't see how it would make a difference for the left-shift operator. Any help and clarification on this will be appreciated. It would be helpful to see the binary representation of the the "result" variable for the ones that I'm asking about. Thanks in advance.

result = (byte) (y << 8);
System.out.println("result: " + result); // 0. Why???
// I was expecting a shift of 0 bits because the
// right-hand operand is equal to the number of
// bits for the size of the result type--in this case
// 8 bits for a byte.the result of (y << 8) is an int, not a byte. the byte "y" is promoted to int for the bit shift. so the int result of the bit shift is 00000000 00000000 00001010 00000000. when you cast that back to byte, the 24 leftmost bits get lopped off, and you're left with zero.
hth,
p

The Shift operator Question

Dear sir/madam
I having question for the shift operator.
What different between >>> and >> ?
I have refer to sun tutorial, it state >>> is unsign, what does it mean?
and last is
if integer is 13
what anwer for 13>>8
is it move for 8 right by position? pls told me in detail
is it answer same with the 13>>>8 ?
Thanks for ur solutions

You can play with the following for a while.
final public class BinaryViewer{
final static char[] coeffs = new char[]{(char)0x30,(char)0x31};
public static String toBinary(int n) {
char[] binary = new char[32];
for(int j=0;j<binary.length;j++) binary[j]=(char)0x30;
int charPointer = binary.length -1;
if(n==0) return "0";
while (n!=0){
      binary[charPointer--] = coeffs[n&1];
      n >>>= 1;
return new String(binary);
public static void print(int n){
    System.out.print(BinaryViewer.toBinary(n));
    System.out.print("   ");
    System.out.println(String.valueOf(n));
public static void main(String[] args) throws Exception{
   int n=Integer.MIN_VALUE;//Integer.parseInt(args[0]);
   int m=Integer.MAX_VALUE;//Integer.parseInt(args[1]);
   BinaryViewer.print(n);
   BinaryViewer.print(n>>8);
   BinaryViewer.print(n>>>8);
   System.out.println();
   BinaryViewer.print(m);
   BinaryViewer.print(m>>8);
   BinaryViewer.print(m>>>8);
System.out.println();
    n=1024;
    m=-1024;
   BinaryViewer.print(n);
   BinaryViewer.print(n>>8);
   BinaryViewer.print(n>>>8);
   System.out.println();
   BinaryViewer.print(m);
   BinaryViewer.print(m>>8);
   BinaryViewer.print(m>>>8);
System.out.println();
   n=2048;
   m=-2048;
   BinaryViewer.print(n);
   BinaryViewer.print(n>>8);
   BinaryViewer.print(n>>>8);
   System.out.println();
   BinaryViewer.print(m);
   BinaryViewer.print(m>>8);
   BinaryViewer.print(m>>>8);
System.out.println();
}

Bitwise Right Shift vs Divide by 2

Something that just pondered across my mind.
Bitwise operators have their uses, and I've always known about right/left shifting to divide/multiply by 2. I recently did some empirical analysis on an algorithm using this technique, and I had to iterate it over a trillion times before I started seeing any timing differences, and even then they were in the order of 0.0001 milliseconds.
Is the performance argument for using bitwise over standard divide still applicable on modern day processors? When I use any kind of algorithms doing this kind of operation, they probably won't loop more than a few million times at most. For readability/understandability I may as well stick with divide by two.
Perhaps it is only significantly faster in certain languages (e.g. C)?
Regards,
Rob.

Yes. You are correct. In modern operators you will not gain anything by using shift operators in place of division. But the point is even though division, multiplication operators are available for you as a high level programmer, in processor level those operations not available. So multiplication and division are carried out using shift operations in the processor level. In hardware level, a processor supports only a few basic operations like addition, negation, shifting. But when you are using java, you are working at a very high layer so that even complex operators are available for you.
Shift operators and other bitwise are very handy if you are manipulation some hardware from within your java application. If you are doing serial port/parallel port programming, you will find these bitwise operators useful.

Shift operator operations

int i=1;
i <<= 31;
i >>= 31;
i >>= 1;
can anyone explain how the result i value=-1
thanks

The high order bit of an integer is the sign. The ">>" operator maintains the sign. See the results of your shifts below which results in a value of -1. Use ">>>" if you do not want to propagate the sign.
i = 1      00000000000000000000000000000001
i <<= 31   10000000000000000000000000000000
i>>= 31    11111111111111111111111111111111
i>>=1      11111111111111111111111111111111

Bitwise shift operator...

I know this is very basic but I've never done this...
If I have an object that is an Integer, and I want to increase it by 1, isn't this how you do it?
Object ob=new Integer(10);
ob=ob>>1;it tells me that >> operator cannot be applied to object, int
So what can it be applied to..?

Ummmm,sadly, no... :(
I really do lack basic computer skills, I don't know
a lot about binary and stuff... but how would you do
what I want to do..?I asked because it's quite a weird code. Just checking.
OK, so:
the binary shift operator can only be applied between to ints, that is primitives, no Objects (like Integer is).
This operator is not used to increase a number, but to shift its bits by one to the right.
Usage:
int i,j;
j = 10;
i = j >> 1;
Integer obj = new Integer(10);
int res = obj.intValue() >> 1;
Integer resObj = new Integer(res);

Can not sign in the operation couldnot be completed

Can not sign in the operation couldnot be completed

Try this...
Go to Settings>icloud, scroll to bottom of screen and tap Delete Account. Then log in again.

TS4036 cannot sign in the operation couldn't becompleted.(com.apple.appleaccount error 403) showing in icloud i need a sloution

dear sir
i can't sing in to my icloud. showing the message as above
cannot sign in the operation couldn't becompleted.(com.apple.appleaccount error 403)
i need a sloution fo it as soon as i am waiting to you ar replay
thank you
mohamed haseef

Welcome to the Apple community.
Are you trying to create a new ID, how many ID's have you created already on that device.

HT5621 i have this error, cannot sign in the operation couldn't be compleated.(com.apple.ppleaccount error 403.)

i have I phone 4s facing this error, cannot sign in the operation couldn't be compleated.(com.apple.ppleaccount error 403.)

Welcome to the Apple community.
Are you trying to create a new ID, how many ID's have you created already on that device.

Wireless keyboard right shift don't work with some keys but left shift works

So, this is really a strange behaviour under Mavericks 10.9.5
My right shift doesn't like some "keys" and some keys don't like the shifts ( either left or right ). For example:
doing a " ? " or a " B " or a " % " with the right shift doesn't work. I mean, after i remove the batteries and stay without using the keyboard for a day or two, when i get back to it, it works for a couple of minutes. But after that... i just can do a capital B or a ? with the right ****. Although i can do all other characters.
the " t " key is another one. No caps here, with either left or right shift key. But the key by itself is working just fine.
So i'm not really 100% convinced it's an hardware fault, because the all keys are working properly in other situations.
Any idea about this strange behaviour?
thank you for your time,
Regards
Pedro.

Hello Postmodum,
Check out the article below and turn on the keyboard viewer to see what happens to the keyboard when you press the right shift key. You may find out that it may be linked to add another combination of keys, which can produce the odd behavior that you are describing.
One or more keys on the keyboard do not respond
http://support.apple.com/kb/TS1381
Regards,
-Norm G.

Spill has caused right shift key to be stuck on

A few months ago I spilled juice on my MBP keyboard (out of warranty - purchased June 2012). I removed all the keys effected and cleaned them. The result of this was that the right shift key was stuck on and it would only start in safe mode + I couldn't log in because I could only type in caps. This is what I did:
I plugged in a usb keyboard - was able to log in
I downloaded keyremapper4mac and disabled the right shift key so I could type in NOT CAPS
Switched to automatic login so I wouldn't be prompted to enter my password anymore
Removed the usb keyboard
Again, MBP would only start in safe mode. Then I found out that the key remapper would only work when the computer wasn't in safe mode. I did some googling and found a key combination that would stop the MBP from starting in safe mode. I have left my computer on for awhile now (not shutting it down) and yesterday I shut it down. I started it up holding down the cmd/shift and my login screen came up.
My problem is that I think I may have forgotten the key combination to bypass the safe mode boot up and I don't know why my login screen came up - maybe because of the key combo I used temporarily turned off the automatic login.
Does anyone know the combination of keys to press to get past the safe boot so my computer will start up normally? I am running 10.7.5.
Thanks!!

Someone else said to try and reboot by holding down the menu and play button on the remote for several seconds and then try again. It worked for several people and seems to be working for me now.

Signed right shift operator

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