How does bitshifting work in Java?

Solution 1:

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

or

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:

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:

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);