Swift - right shift assignment operator

The Bitwise right shift assignment operator (>>=) assigns the first operand a value equal to the result of Bitwise right shift operation of two operands.

(x >>= y) is equivalent to (x = x >> y)

The Bitwise right shift operator (>>) takes the two numbers and right shift the bits of first operand by number of place specified by second operand. For example: for right shifting the bits of x by y places, the expression (x>>y) can be used. It is equivalent to dividing x by 2^y.

The example below describes how right shift operator works:

1000 >> 2 returns 250

                      (In Binary)
   1000         ->    1111101000  
   >> 2                     |  right shift the bits
   -----                    V  by 2 places
    250         <-      11111010 
                      (In Binary) 

The code of using right shift operator (>>) is given below:

var x = 1000

//right shift assignment operation
x >>= 2

//Displaying the result
print("x = \(x)")

The output of the above code will be:

x = 250

Example: Find largest power of 2 less than or equal to given number

Consider an integer 1000. In the bit-wise format, it can be written as 1111101000. However, all bits are not written here. A complete representation will be 32 bit representation as given below:

00000000000000000000001111101000  

Performing n = n | (n>>i) operation, where i = 1, 2, 4, 8, 16 will change all right side bit to 1. When applied on 1000, the result in 32 bit representation is given below:

00000000000000000000001111111111 

Adding one to this result and then right shifting the result by one place will give largest power of 2 less than or equal to 1000.

00000000000000000000001000000000 

The below code will calculate the largest power of 2 less than or equal to given number.

func MaxPowerOfTwo(_ n: Int) -> Int {
  var n = n
  //changing all right side bits to 1.
  n = n | (n>>1)
  n = n | (n>>2)
  n = n | (n>>4)
  n = n | (n>>8)
  n = n | (n>>16)
  
  //adding 1 to n makes smallest power
  //of 2 greater than given number
  n = n + 1

  //right shift by one position makes
  //largest power of 2 less than or 
  //equal to given number
  n >>= 1
  
  return n
}

print("MaxPowerOfTwo(100) = \(MaxPowerOfTwo(100))")
print("MaxPowerOfTwo(500) = \(MaxPowerOfTwo(500))")
print("MaxPowerOfTwo(1000) = \(MaxPowerOfTwo(1000))")  

The above code will give the following output:

MaxPowerOfTwo(100) = 64
MaxPowerOfTwo(500) = 256
MaxPowerOfTwo(1000) = 512