Conversion of decimal into binary using recursion

Question

I want to convert the decimal number into a binary number using recursion in java. I tried a lot but unable to do it. Here is my code:

public class DecimalToBinary {


    public static void main(String[] args) {
        System.out.println(conversion(2));
    }

    public static int conversion(int n) {
         return reconversion(n);
    }

    public static int reconversion(int n) {
        if(n <= 0)
            return 0;
        else {

            return  (int) (n/2 + conversion(n/2));

        }
    }

}

WJS · Accepted Answer · 2019-10-19 17:44:53Z

2

Integer values are already in binary. The fact that they appear as digits 0 thru 9 when you print them is because they are converted to a string of decimal digits. So you need to return a String of binary digits like so.

   public static String conversion(int n) {
      String b = "";
      if (n > 1) {
         // continue shifting until n == 1
         b = conversion(n >> 1);
      }
      // now concatenate the return values based on the logical AND
      b += (n & 1);
      return b;

   }

answered Oct 19, 2019 at 17:44

WJS

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3 Comments

WJS Over a year ago

It shifts the value of n one bit to the right. So if n is 101 in binary, it will now call conversion with 10.

Yahya Over a year ago

And what about the statement b+ = (n & 1); ?? I am unaware of ( n & 1). Kindly help me

WJS Over a year ago

That is a bit test for a value. It does a logical AND operation with n and a give mask. Since the mask is 1, it checks to see if the low order bit is set. It if is (1 & 1) returns a 1. If it isn't (0 & 1) returns a 0. Other bit operations are ~ and ^ and | which are complement, exclusive OR, and OR.

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