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

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);

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.

Search for "Java literals syntax" on Google and you come up with some entries.

There is an octal syntax (prefix your number with 0), decimal syntax and hexadecimal syntax with a "0x" prefix. But no syntax for binary notation.

Some examples:

int i = 0xcafe ; // hexadecimal case
int j = 045 ;    // octal case
int l = 42 ;     // decimal case

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"