How to get 0-padded binary representation of an integer in java?

Question

for example, for 1, 2, 128, 256 the output can be (16 digits):

0000000000000001
0000000000000010
0000000010000000
0000000100000000

Answer 1

A simpler version of user3608934's idea "This is an old trick, create a string with 16 0's then append the trimmed binary string you got ":

private String toBinaryString32(int i) {
    String binaryWithOutLeading0 = Integer.toBinaryString(i);
    return "00000000000000000000000000000000"
            .substring(binaryWithOutLeading0.length())
            + binaryWithOutLeading0;
}

Answer 2

I think this is a suboptimal solution, but you could do

String.format("%16s", Integer.toBinaryString(1)).replace(' ', '0')

Answer 3

This method converts an int to a String, length=bits. Either padded with 0s or with the most significant bits truncated.

static String toBitString( int x, int bits ){
    String bitString = Integer.toBinaryString(x);
    int size = bitString.length();
    StringBuilder sb = new StringBuilder( bits );
    if( bits > size ){
        for( int i=0; i<bits-size; i++ )
            sb.append('0');
        sb.append( bitString );
    }else
        sb = sb.append( bitString.substring(size-bits, size) );

    return sb.toString();
}

Answer 4

for(int i=0;i<n;i++)
{
  for(int j=str[i].length();j<4;j++)
  str[i]="0".concat(str[i]);
}

str[i].length() is length of number say 2 in binary is 01 which is length 2 change 4 to desired max length of number. This can be optimized to O(n). by using continue.

Answer 5

// Below will handle proper sizes

public static String binaryString(int i) {
    return String.format("%" + Integer.SIZE + "s", Integer.toBinaryString(i)).replace(' ', '0');
}

public static String binaryString(long i) {
    return String.format("%" + Long.SIZE + "s", Long.toBinaryString(i)).replace(' ', '0');
}

Answer 6

import java.util.Scanner;
public class Q3{
  public static void main(String[] args) {
    Scanner scn=new Scanner(System.in);
    System.out.println("Enter a number:");
    int num=scn.nextInt();
    int numB=Integer.parseInt(Integer.toBinaryString(num));
    String strB=String.format("%08d",numB);//makes a 8 character code
    if(num>=1 && num<=255){
     System.out.println(strB);
    }else{
        System.out.println("Number should be in range between 1 and 255");
    }
  }
}