How does bitshifting work in Java?

Java Problem Overview

I have this statement:

> Assume the bit value of byte x is 00101011. what is the result of x>>2?

How can I program it and can someone explain me what is doing?

Java Solutions

Solution 1 - Java

Firstly, you can not shift a byte in java, you can only shift an int or a long. So the byte will undergo promotion first, e.g.

00101011 -> 00000000000000000000000000101011

11010100 -> 11111111111111111111111111010100

Now, x >> N means (if you view it as a string of binary digits):

The rightmost N bits are discarded
The leftmost bit is replicated as many times as necessary to pad the result to the original size (32 or 64 bits), e.g.

00000000000000000000000000101011 >> 2 -> 00000000000000000000000000001010

11111111111111111111111111010100 >> 2 -> 11111111111111111111111111110101

Solution 2 - Java

Shift Operators

The binary 32 bits for 00101011 is

00000000 00000000 00000000 00101011, and the result is:

  00000000 00000000 00000000 00101011   >> 2(times)
 \\                                 \\
  00000000 00000000 00000000 00001010

Shifts the bits of 43 to right by distance 2; fills with highest(sign) bit on the left side.

Result is 00001010 with decimal value 10.

00001010
    8+2 = 10

Solution 3 - Java

When you shift right 2 bits you drop the 2 least significant bits. So:

x = 00101011

x >> 2

// now (notice the 2 new 0's on the left of the byte)
x = 00001010

This is essentially the same thing as dividing an int by 2, 2 times.

In Java

byte b = (byte) 16;
b = b >> 2;
// prints 4
System.out.println(b);

Solution 4 - Java

These examples cover the three types of shifts applied to both a positive and a negative number:

// Signed left shift on 626348975
00100101010101010101001110101111 is   626348975
01001010101010101010011101011110 is  1252697950 after << 1
10010101010101010100111010111100 is -1789571396 after << 2
00101010101010101001110101111000 is   715824504 after << 3

// Signed left shift on -552270512
11011111000101010000010101010000 is  -552270512
10111110001010100000101010100000 is -1104541024 after << 1
01111100010101000001010101000000 is  2085885248 after << 2
11111000101010000010101010000000 is  -123196800 after << 3


// Signed right shift on 626348975
00100101010101010101001110101111 is   626348975
00010010101010101010100111010111 is   313174487 after >> 1
00001001010101010101010011101011 is   156587243 after >> 2
00000100101010101010101001110101 is    78293621 after >> 3

// Signed right shift on -552270512
11011111000101010000010101010000 is  -552270512
11101111100010101000001010101000 is  -276135256 after >> 1
11110111110001010100000101010100 is  -138067628 after >> 2
11111011111000101010000010101010 is   -69033814 after >> 3


// Unsigned right shift on 626348975
00100101010101010101001110101111 is   626348975
00010010101010101010100111010111 is   313174487 after >>> 1
00001001010101010101010011101011 is   156587243 after >>> 2
00000100101010101010101001110101 is    78293621 after >>> 3

// Unsigned right shift on -552270512
11011111000101010000010101010000 is  -552270512
01101111100010101000001010101000 is  1871348392 after >>> 1
00110111110001010100000101010100 is   935674196 after >>> 2
00011011111000101010000010101010 is   467837098 after >>> 3

Solution 5 - Java

>> is the Arithmetic Right Shift operator. All of the bits in the first operand are shifted the number of places indicated by the second operand. The leftmost bits in the result are set to the same value as the leftmost bit in the original number. (This is so that negative numbers remain negative.)

Here's your specific case:

00101011
  001010 <-- Shifted twice to the right (rightmost bits dropped)
00001010 <-- Leftmost bits filled with 0s (to match leftmost bit in original number)

Solution 6 - Java

public class Shift {
 public static void main(String[] args) {
  Byte b = Byte.parseByte("00101011",2);
  System.out.println(b);
  byte val = b.byteValue();
  Byte shifted = new Byte((byte) (val >> 2));
  System.out.println(shifted);

  // often overloked  are the methods of Integer
  
  int i = Integer.parseInt("00101011",2);
  System.out.println( Integer.toBinaryString(i));
  i >>= 2;
  System.out.println( Integer.toBinaryString(i));
 }
}

Output:

Solution 7 - Java

You can't write binary literals like 00101011 in Java so you can write it in hexadecimal instead:

byte x = 0x2b;

To calculate the result of x >> 2 you can then just write exactly that and print the result.

System.out.println(x >> 2);

Solution 8 - Java

byte x = 51; //00101011
byte y = (byte) (x >> 2); //00001010 aka Base(10) 10

Solution 9 - Java

You can use e.g. this API if you would like to see bitString presentation of your numbers. Uncommons Math

Example (in jruby)

bitString = org.uncommons.maths.binary.BitString.new(java.math.BigInteger.new("12").toString(2))
bitString.setBit(1, true)
bitString.toNumber => 14

edit: Changed api link and add a little example

Solution 10 - Java

00101011 = 43 in decimal

class test {	
    public static void main(String[] args){
	   int a= 43;		
	   String b= Integer.toBinaryString(a >> 2);		
	   System.out.println(b);
    }	
}