How to get 0-padded binary representation of an in

Question

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

0000000000000001
0000000000000010
0000000010000000
0000000100000000

I tried

String.format("%16s", Integer.toBinaryString(1));

it puts spaces for left-padding:

`               1'

How to put 0s for padding. I couldn't find it in Formatter. Is there another way to do it?

Thanks in advance.

P.S. this post describes how to format integers with left 0-padding, but it is not for the binary representation.

Answer 1

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 2

You can use lib https://github.com/kssource/BitSequence. It accept a number and return bynary string, padded and/or grouped.

String s = new BitSequence(2, 16).toBynaryString(ALIGN.RIGHT, GROUP.CONTINOUSLY));  
return  
0000000000000010  

another examples:

[10, -20, 30]->00001010 11101100 00011110
i=-10->00000000000000000000000000001010
bi=10->1010
sh=10->00 0000 0000 1010
l=10->00000001 010
by=-10->1010
i=-10->bc->11111111 11111111 11111111 11110110

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

This is an old trick, create a string with 16 0's then append the trimmed binary string you got from String.format("%s", Integer.toBinaryString(1)) and use the right-most 16 characters, lopping off any leading 0's. Better yet, make a function that lets you specify how long of a binary string you want. Of course there are probably a bazillion other ways to accomplish this including libraries, but I'm adding this post to help out a friend :)

public class BinaryPrinter {

    public static void main(String[] args) {
        System.out.format("%d in binary is %s\n", 1, binaryString(1, 4));
        System.out.format("%d in binary is %s\n", 128, binaryString(128, 8));
        System.out.format("%d in binary is %s\n", 256, binaryString(256, 16));
    }

    public static String binaryString( final int number, final int binaryDigits ) {
        final String pattern = String.format( "%%0%dd", binaryDigits );
        final String padding = String.format( pattern, 0 );
        final String response = String.format( "%s%s", padding, Integer.toBinaryString(number) );

        System.out.format( "\npattern = '%s'\npadding = '%s'\nresponse = '%s'\n\n", pattern, padding, response );

        return response.substring( response.length() - binaryDigits );
    }
}

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

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

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