Java - Convert decimal to binary - Integer - Recursive function

Question

Hey I'm really trying to convert decimal int to binary int, without succes.

Please help me.

And I don't want to make it 'System.out.println()' becuase I've already done it.

THANKS!

`I need recursive function that gets decimal int and return binary int

public static void decToBin(int number) {
    if(number == 0)
        return ;
    decToBin(number / 2);
    System.out.print(number % 2);
}

That it what I've done...

When I'm trying to get string:

public static String decToBin(int number) {
    if(number == 0)
        return "";
    return new Integer(number % 2).toString() + new Integer(decToBin(number / 2)).toString();
}

Error...

Answer 1

public static String decToBin(int number) {
    if(number == 0)
        return "";
    return new Integer(number % 2).toString() + new Integer(decToBin(number / 2)).toString();
}

This approach has few flaws.

• First you can't use "" as argument of new Integer("") because "" holds no value, so you will see NumberFormatException.

• Other problem is order of generating result. number % 2 should be placed after result of decToBin(number / 2), just like you are doing it in your first example, where you are printing it after recursive decToBin call

  decToBin(number / 2);
  System.out.print(number % 2);
  

• Lastly in your return statement you are creating new string, so to generate "100101" you will generate "" "1" "01" "101" 0101" "00101" and then finally "100101". To avoid it use StringBuilder and its append method.

Answer 2

This is a little old, but for the sake of reference and giving a correct answer, this is what you would need to do:

public static String dec2Bin(int num) {

    String result = ((num % 2 == 0) ? "0" : "1");

    if (num == 0 || num == 1) {
        return result;
    }

    return dec2Bin(num/2) + result;
}

Answer 3

    void intToBinary(int num) {

    if (num < 2) {
        System.out.print(num);
        return;
    } else {
        intToBinary(num / 2);
        intToBinary(num % 2);
    }

}

Answer 4

import java.io.*;

class Binary{
    public static void main(String args[])throws IOException{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        System.out.print("Enter the number:");
        int n = Integer.parseInt(br.readLine());
        System.out.print(Integer.toBinaryString(n));
    }
}