Binary representation in Java

Question

I am finding it difficult to understand and work with this binary representation in java:

With the help of the user Jon Skeet, I understood that binary representation should be built this way.

Here's a code sample:

public class chack {

public static void main(String[] args) {
    int num2=2;
    int num3=3;
    int num4=4;
    int num1=1;
    int nirbinary = (num1 << 24) | (num2 << 16) | (num3 << 8) | num4;
    System.out.println(nirbinary);
    String nir=  Integer.toBinaryString(nirbinary);
    System.out.println(nir);
    }
}

Couple of question:

1. How does one get num1 (for example) back from an int who is already in this binary
2. why do I get 16909060 when I print nirbinary- what does it stands for? How does one get num1 (for example) back from an int who is already in this binary representation?

Thank you

Answer 1

16909060 stands for the number 16909060.

It is (1 * 2²⁴) + (2 * 2¹⁶) + (3 * 2⁸) + 4.

To get num1 back out, just right-shift the result the same amount you left-shifted and mask out the other bytes (not always necessary for num1^(*), but for the others):

int num1 = nirbinary >> 24 & 0xFF;
int num2 = nirbinary >> 16 & 0xFF;
int num3 = nirbinary >> 8 & 0xFF;
int num4 = nirbinary & 0xFF;

Note that nirbinary is not "a binary representation". Or more precisely: it's no more or less binary than num1, num2, num3 and num4: internally all numbers (and characters, and booleans, ...) are stored in binary.

(*) note that if num1 is > 127, then you either need to use >>> to do the right-shift or use the & 0xFF in order to ensure that the correct value is restored. The difference between >> and >>> are the "new" bits inserted on the "left" side of the value: With >> they will depend on the highest-value bit (known as sign-extension) and with >>> they will always be 0.

Answer 2

I am not completely sure what you are missing, so I will just explain how you can convert integers to binary strings back and forth in java.

You can get a binary string from an integer like so:

int i = 1234;
String binString = Integer.toBinaryString(i);

and you can convert the string back to an integer this way:

int iNew = Integer.parseInt(binString, 2);

Note the second argument to Integer.parseInt() is the desired base of the number. 2 is binary, 8 is octal, 10 decimal, etc.