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

You can also use

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

EDIT: the difference between "(value>>x) & 1" and "value & (1<" 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<" (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<" are completely equivalent