Java: Checking if a bit is 0 or 1 in a long

Question

What method would you use to determine if the the bit that represents 2^x is a 1 or 0 ?

Answer 1

Another alternative:

if (BigInteger.valueOf(value).testBit(x)) {
    // ...
}

Answer 2

I wonder if:

  if (((value >>> x) & 1) != 0) {

  }

.. is better because it doesn't matter whether value is long or not, or if its worse because it's less obvious.

Tom Hawtin - tackline Jul 7 at 14:16

Answer 3

In Java the following works fine:

if (value << ~x < 0) {
   // xth bit set
} else {
   // xth bit not set
}

value and x can be int or long (and don't need to be the same).

Word of caution for non-Java programmers: the preceding expression works in Java because in that language the bit shift operators apply only to the 5 (or 6, in case of long) lowest bits of the right hand side operand. This implicitly translates the expression to value << (~x & 31) (or value << (~x & 63) if value is long).

Javascript: it also works in javascript (like java, only the lowest 5 bits of shift count are applied). In javascript any number is 32-bit.

Particularly in C, negative shift count invokes undefined behavior, so this test won't necessarily work (though it may, depending on your particular combination of compiler/processor).

Answer 4

Bit shifting right by x and checking the lowest bit.

Answer 5

You can also use

bool isSet = ((value>>x) & 1) != 0;

EDIT: the difference between "(value>>x) & 1" and "value & (1<<x)" relies on the behavior when x is greater than the size of the type of "value" (32 in your case).

In that particular case, with "(value>>x) & 1" you will have the sign of value, whereas you get a 0 with "value & (1<<x)" (it is sometimes useful to get the bit sign if x is too large).

If you prefer to have a 0 in that case, you can use the ">>>" operator, instead if ">>"

So, "((value>>>x) & 1) != 0" and "(value & (1<<x)) != 0" are completely equivalent

Answer 6

You might want to check out BitSet: http://java.sun.com/javase/6/docs/api/java/util/BitSet.html