Converting Decimal to Binary Java

Your binaryForm method is getting caught in an infinite recursion, you need to return if number <= 1:

import java.util.Scanner;

public class ReversedBinary {

    public static void main(String[] args) {
        int number;

        Scanner in = new Scanner(System.in);

        System.out.println("Enter a positive integer");
        number = in.nextInt();

        if (number < 0) {
            System.out.println("Error: Not a positive integer");
        } else {

            System.out.print("Convert to binary is:");
            //System.out.print(binaryform(number));
            printBinaryform(number);
        }
    }

    private static void printBinaryform(int number) {
        int remainder;

        if (number <= 1) {
            System.out.print(number);
            return; // KICK OUT OF THE RECURSION
        }

        remainder = number % 2;
        printBinaryform(number >> 1);
        System.out.print(remainder);
    }
}

Integer.toString(n,8) // decimal to octal

Integer.toString(n,2) // decimal to binary

Integer.toString(n,16) //decimal to Hex

where n = decimal number.

I just want to add, for anyone who uses:

   String x=Integer.toBinaryString()

to get a String of Binary numbers and wants to convert that string into an int. If you use

  int y=Integer.parseInt(x)

you will get a NumberFormatException error.

What I did to convert String x to Integers, was first converted each individual Char in the String x to a single Char in a for loop.

  char t = (x.charAt(z));

I then converted each Char back into an individual String,

  String u=String.valueOf(t);

then Parsed each String into an Integer.

Id figure Id post this, because I took me a while to figure out how to get a binary such as 01010101 into Integer form.

Integer.toBinaryString() is an in-built method and will do quite well.