How to convert a byte to its binary string representation

How to convert a byte to its binary string representation

For example, the bits in a byte B are 10000010, how can I assign the bits to the string strliterally, that is, str = "10000010".

byte b1 = (byte) ;
String s1 = String.format("%8s", Integer.toBinaryString(b1 & 0xFF)).replace(' ', '');
System.out.println(s1); //

byte b2 = (byte) ;
String s2 = String.format("%8s", Integer.toBinaryString(b2 & 0xFF)).replace(' ', '');
System.out.println(s2); //

public class BitsSetCount
{
    public static void main(String[] args)
    {
        int send = ;

        System.out.println( "[Input] Integer value: " + send + "\n" );
        BitsSetCount.countBits(  send );
    }

    private static void countBits(int i)
    {
        System.out.println( "Integer.toBinaryString: " + Integer.toBinaryString(i) );
        System.out.println( "Integer.toHexString: " + Integer.toHexString(i) );
        System.out.println( "Integer.bitCount: "+ Integer.bitCount(i) );

        int d = i & 0xff000000;
        int c = i & 0xff0000;
        int b = i & 0xff00;
        int a = i & 0xff;

        System.out.println( "\nByte 4th Hex Str: " + Integer.toHexString(d) );
        System.out.println( "Byte 3rd Hex Str: " + Integer.toHexString(c) );
        System.out.println( "Byte 2nd Hex Str: " + Integer.toHexString(b) );
        System.out.println( "Byte 1st Hex Str: " + Integer.toHexString(a) );

        int all = a+b+c+d;
        System.out.println( "\n(1st + 2nd + 3rd + 4th (int(s)) as Integer.toHexString: " + Integer.toHexString(all) );

        System.out.println("(1st + 2nd + 3rd + 4th (int(s)) ==  Integer.toHexString): " +
                Integer.toHexString(all).equals(Integer.toHexString(i) ) );

        System.out.println( "\nIndividual bits for each byte in a 4 byte int:");

        /*
         * Because we are sending the MSF bytes to a method
         * which will work on a single byte and print some
         * bits we are generalising the MSF bytes
         * by making them all the same in terms of their position
         * purely for the purpose of printing or analysis
         */
        System.out.print(
                    getBits( (byte) (d >> ) ) + " " +
                    getBits( (byte) (c >> ) ) + " " +
                    getBits( (byte) (b >> ) ) + " " +
                    getBits( (byte) (a >> ) )
        );

    }

    private static String getBits( byte inByte )
    {
        // Go through each bit with a mask
        StringBuilder builder = new StringBuilder();
        for ( int j = ; j < ; j++ )
        {
            // Shift each bit by 1 starting at zero shift
            byte tmp =  (byte) ( inByte >> j );

            // Check byte with mask 00000001 for LSB
            int expect1 = tmp & 0x01; 

            builder.append(expect1);
        }
        return ( builder.reverse().toString() );
    }

}

public static String byteToString(byte b) {
    byte[] masks = { -, , , , , , ,  };
    StringBuilder builder = new StringBuilder();
    for (byte m : masks) {
        if ((b & m) == m) {
            builder.append('');
        } else {
            builder.append('');
        }
    }
    return builder.toString();
}

public static String getByteBinaryString(byte b) {
    StringBuilder sb = new StringBuilder();
    for (int i = ; i >= ; --i) {
        sb.append(b >>> i & );
    }
    return sb.toString();
}

String byteToBinaryString(byte b){
    StringBuilder binaryStringBuilder = new StringBuilder();
    for(int i = ; i < ; i++)
        binaryStringBuilder.append(((0x80 >>> i) & b) == ? '':'');
    return binaryStringBuilder.toString();
}