Java の符号なしおよび符号付き右ビット シフト演算子

Let's take a number -11, and we will shift 2 bits towards the right.

+11 in 8-bit form = 0000 1011
1's Complement    = 1111 0100
2's Compelement   =        +1
-----------------------------
2's Complement of -11 = 1111 0101

n = -11 =  1111 0101
Shift 1st bit = 1111 1010
Shift 2nd bit = 1111 1101

1111 1101
       -1
------------------------------
1111 1100 = 1's Complement
0000 0011 = Complement of each bit

So -11 >> 2 = -3 which is in binary 0000 0011.

n = 10 = 0000 1010 //the leftmost bit position is filled with '0' as 'n' is positive

Now Shift the '3' bits towards the right; you can do it directly, but let's go one-by-one
n = 0000 1010
Shift 1st bit = 0000 0101
Shift 2nd bit = 0000 0010
Shift 3rd bit = 0000 0001

So, n >>> 3 = 0000 0001 which would be 1 in decimals.

We can also use this formula:(Given decimal number/2^n)
Where 'n' is the required number of shifts, in our case, it is '3'.

So, n >>> 3 = 10/2^3 = 10/8 = 1

public class SignedRightShift {
  public static void main(String args[]) {
    int a = -11;
    System.out.println(a >> 2);

    int b = 4;
    System.out.println(b >> 1);
  }
}

-3
2

public class UnsignedRightShift {
  public static void main(String args[]) {
    // The binary representation of -1 is all "111..1".
    int a = -1;

    // The binary value of 'a >>> 29' is "00...0111"
    System.out.println(a >>> 29);

    // The binary value of 'a >>> 30' is "00...0011"
    System.out.println(a >>> 30);

    // The value of 'a >>> 31' is "00...0001"
    System.out.println(a >>> 31);
  }
}

7
3
1