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Scala - 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 2y.

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:

object MainObject {
  def main(args: Array[String]) {
    var x = 1000

    //right shift assignment operation
    x >>= 2

    //Displaying the result
    println("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.

object MainObject {
  def 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
  }

  def main(args: Array[String]) {
    println(s"MaxPowerOfTwo(100) = ${MaxPowerOfTwo(100)}")
    println(s"MaxPowerOfTwo(500) = ${MaxPowerOfTwo(500)}")
    println(s"MaxPowerOfTwo(1000) = ${MaxPowerOfTwo(1000)}") 
  }
}  

The above code will give the following output:

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

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