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

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).