In Java, can I define an integer constant in binary format?

Question

Similar to how you can define an integer constant in hexadecimal or octal, can I do it in binary?

I admit this is a really easy (and stupid) question. My google sea

Answer 1

If you want to mess around with lots of binary you could define some constants:

public static final int BIT_0 = 0x00000001;
public static final int BIT_1 = 0x00000002;

etc.

or

public static final int B_00000001 = 0x00000001;
public static final int B_00000010 = 0x00000002;
public static final int B_00000100 = 0x00000004;

Answer 2

Using binary constants to masking

Declare constants:

public static final int FLAG_A = 1 << 0;
public static final int FLAG_B = 1 << 1;
public static final int FLAG_C = 1 << 2;
public static final int FLAG_D = 1 << 3;

and use them

if( (value & ( FLAG_B | FLAG_D )) != 0){
    // value has set FLAG_B and FLAG_D
}

Answer 3

The answer from Ed Swangren

public final static long mask12 = 
  Long.parseLong("00000000000000000000100000000000", 2);

works fine. I used long instead of int and added the modifiers to clarify possible usage as a bit mask. There are, though, two inconveniences with this approach.

1. The direct typing of all those zeroes is error prone
2. The result is not available in decimal or hex format at the time of development

I can suggest alternative approach

public final static long mask12 = 1L << 12;

This expression makes it obvious that the 12th bit is 1 (the count starts from 0, from the right to the left); and when you hover mouse cursor, the tooltip

long YourClassName.mask12 = 4096 [0x1000]

appears in Eclipse. You can define more complicated constants like:

public final static long maskForSomething = mask12 | mask3 | mask0;

or explicitly

public final static long maskForSomething = (1L<<12)|(1L<<3)|(1L<<0);

The value of the variable maskForSomething will still be available in Eclipse at development time.

Answer 4

So, with the release of Java SE 7, binary notation comes standard out of the box. The syntax is quite straight forward and obvious if you have a decent understanding of binary:

byte fourTimesThree = 0b1100;
byte data = 0b0000110011;
short number = 0b111111111111111; 
int overflow = 0b10101010101010101010101010101011;
long bow = 0b101010101010101010101010101010111L;

And specifically on the point of declaring class level variables as binaries, there's absolutely no problem initializing a static variable using binary notation either:

public static final int thingy = 0b0101;

Just be careful not to overflow the numbers with too much data, or else you'll get a compiler error:

byte data = 0b1100110011; // Type mismatch: cannot convert from int to byte

Now, if you really want to get fancy, you can combine that other neat new feature in Java 7 known as numeric literals with underscores. Take a look at these fancy examples of binary notation with literal underscores:

int overflow = 0b1010_1010_1010_1010_1010_1010_1010_1011;
long bow = 0b1__01010101__01010101__01010101__01010111L;

Now isn't that nice and clean, not to mention highly readable?

I pulled these code snippets from a little article I wrote about the topic over at TheServerSide. Feel free to check it out for more details:

Java 7 and Binary Notation: Mastering the OCP Java Programmer (OCPJP) Exam

Answer 5

There are no binary literals in Java, but I suppose that you could do this (though I don't see the point):

int a = Integer.parseInt("10101010", 2);

Answer 6

Slightly more awkward answer:

public class Main {

    public static void main(String[] args) {
        byte b = Byte.parseByte("10", 2);
        Byte bb = new Byte(b);
        System.out.println("bb should be 2, value is \"" + bb.intValue() + "\"" );
    }

}

which outputs [java] bb should be 2, value is "2"