What's the error here?

Question

int i = 10000;
Integer.toBinaryString(i);

How can I make the Integer.toBinaryString method return leading zeroes as well? For example, for i = 1000, I want 00000000000000000000001111101000 to appear, not 1111101000.

• Freedi: And yes, i did read there's other ways by doing it, but i dont want to copy n paste as i thought this would do the job
• chRyNaN: From the documentation: This value is converted to a string of ASCII digits in binary (base 2) with no extra leading 0s. docs.oracle.com/javase/7/docs/api/java/lang/…
• Freedi: But whats up with the wierd result? What am i doing wrong?
• chRyNaN: The result is not weird it's binary (you know, one's and zero's). If you look at a Decimal to Binary Converter you will see that it is correct just without the leading zero's binaryhexconverter.com/decimal-to-binary-converter. If you just want the int in a String representation of it's int value, use Integer.toString()
• Freedi: Okay sorry it's because i was talking about 8 bits only, sorry, and thanks for help! :)

Answer 1

If you want to left pad the result with zeros, you could do:

String raw = Integer.toBinaryString(i);
String padded = "0000000000000000".substring(raw.length()) + raw;

Here I chose a width of 16 digits, you can adjust the width by the number of zeros in the string.

Note, if it is possible that i > 2^16 - 1 then this will fail and you'll need to protect against that (32 zeros would be one approach).

EDIT

Here's a more complicated version which formats to the smallest of 8, 16, 24, or 32 bits which will contain the result:

public class pad {
 public static String pbi ( int i ) {
 String raw = Integer.toBinaryString(i);
 int n = raw.length();
 String zeros;
 switch ((n-1)/8) {
 case 0: zeros = "00000000"; break;
 case 1: zeros = "0000000000000000"; break;
 case 2: zeros = "000000000000000000000000"; break;
 case 3: zeros = "00000000000000000000000000000000"; break;
 default: return raw;
 }
 return zeros.substring(n) + raw;
 }
 public static void main ( String[] args ) {
 Scanner s = new Scanner(System.in);
 System.out.print("Enter an integer : ");
 int i = s.nextInt();
 System.out.println( pbi( i ) );
 }
}

• Freedi: Wouldn't this throw a error if its over 16 digits??
• John Hascall: Yup, if that's a possibility, you'll need to deal with it somehow.