CODEWITHSUNDEEP
javabit-manipulationbytebit
Jesus
Jesus

Reputation: 3

Issue finding out bits values from byte in Java

I am having an issue trying to find the bit-wise values of a Java byte type. When I try to obtain the value of a certain number of bits by using the bit-wise operator & this value is not the way I foresee it. The (very simple) code is as follows:

public class Demo 
{
    public static void main(String[]arg)
    {
        byte demo = 127;
        System.out.println("demo is: "+(demo & 00000011));
    }
}

In this case, I expect to obtain a 3 (since what I am expecting is 01111111 AND 00000011 = 00000011) but I obtain a 9 (00001001). If I do it with 00000111 then the result is 73 (01001001) rather than 7. Why are there two zeros that are introduced in the comparison?

I have checked the answers that were given to other questions but they are not working in my case, or are not exactly the same thing i am requesting. I am sure there must be some kind of minor thing but I am unable to find how.

Thanks in advance

Upvotes: 0

Views: 91

Answers (2)

user1933888
user1933888

Reputation: 3017

You can directly do bitwise operators on int values (Java will do the binary conversion)

public static void main(String[] args) throws ParseException {
        int demo = 127;
        System.out.println(demo);
        System.out.println("demo is: " + (demo & 3));
    }

Reference:http://www.tutorialspoint.com/java/java_bitwise_operators_examples.htm https://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html

or you can convert your octal value to binary representation as:

00000011
to 
0b00000011

Which is already pointed in above answer

Upvotes: 0

user6073886
user6073886

Reputation:

As Patricia already pointed out: Java interpretes numbers starteing with leading zeros as being octal values.

If you want the number to be binary, mark it with a leading "0b":

System.out.println("demo is: "+(demo & 0b00000011));

Upvotes: 3

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