Suppose x is a bitmask value, and b is one flag, e.g.
x = 0b10101101
b = 0b00000100
There seems to be two ways to check whether the bit indicated by b is turned on in x:
if (x & b != 0) // (1)
if (x & b == b) // (2)
In most circumstances it seems these two checks always yield the same result, given that b is always a binary with only one bit turned on.
However I wonder is there any exception that makes one method better than another?
In general, if we interpret both values as bit sets, the first condition checks if the intersection of x and b is not empty (or, to put it differently: if b and x have elements in common), while the second one checks if b is a subset of x.
Clearly, if b is a singleton, b is a subset of x if and only if the intersection is not empty.
So, whenever you cannot guarantee to 100% that b is a singleton, choose your condition wisely. Ask yourself if you want to express that all elements of b must also be elements of x, or that there are elements of b that are also elements of x. It's a huge difference except for the single bit case.
