Implement Multiplication of two unsigned n-bit binary numbers with shift and add method. > Java Program

Computer Organization and Architecture

A Java Program for Implement Multiplication of two unsigned n-bit binary numbers with shift and add method. The numbers are taken as input from the user in decimal form and must be converted from decimal to binary form as required. The final answer should be shown to the user in both decimal and binary form.

import java.util.*; //Importing all classes of Util Package
public class BitMultiply //Creating a class named BitMultiply
 {

 public static void main(String[] args) //Main Method
  {
  System.out.println("Enetr the numbers in decimal form which you want to multiply");
  Scanner InpOb = new Scanner(System.in); //Making Object InpOb
  int num1 = InpOb.nextInt();
  int num2 = InpOb.nextInt();
  int i1=0, i2=0, i3=0;
         int arr1[] = new int[100];
         int arr2[] = new int[100];
         int arr3[] = new int[100];
  int a = num1;
  System.out.println(num1 + " in binary form is ");
  while (a != 0)      //Converting num1 in Binary
          {
          i1++;
          arr1[i1] = a % 2;
              a = a / 2;
          }
         for (int j = i1; j > 0; j--)
          {
          System.out.print(arr1[j]);
          }
  System.out.println();
  a = num2;
  System.out.println(num2 + " in binary form is ");
  while (a != 0)       //Converting num2 in Binary
          {
          i2++;
          arr2[i2] = a % 2;
              a = a / 2;
          }
         for (int j = i2; j > 0; j--)
          {
          System.out.print(arr2[j]);
          }
  System.out.println();
  System.out.println("Multiplying " +num1 +" & " + num2 +" in we get");
  int result = 0;
  while (num2 != 0) // Iterate the loop till b==0
          {
              if ((num2 & 01) != 0) // Logical ANDing of the value of num2 with 01
              {
                  result = result + num1; // Update the result with the new value of num1.
              }
              num1 <<= 1;              // Left shifting the value contained in 'num1' by 1.
              num2 >>= 1;             // Right shifting the value contained in 'num2' by 1.
          }
          System.out.println(result);
  System.out.println();
  a = result;
  System.out.println(result + " in binary form is ");
  while (a != 0)
          {
          i3++;
          arr3[i3] = a % 2;
              a = a / 2;
          }
         for (int j = i3; j > 0; j--)
          {
          System.out.print(arr3[j]);
          }

  }
 }

/*Result
Enetr the numbers in decimal form which you want to multiply
23
54
23 in binary form is
10111
54 in binary form is
110110
Multiplying 23 & 54 in we get
1242

1242 in binary form is
1001101101
*/