Hex combination of binary flags

Question

Which of the following give back 63 as long (in Java) and how?

0x0
0x1
0x2
0x4
0x8
0x10
0x20

I'm working with NetworkManager API flags if that helps. I'm getting 63 from one of the operations but don't know how should I match the return value to the description.

Thanks

Answer 1

63 is 32 | 16 | 8 | 4 | 2 | 1, where | is the binary or operator.

Or in other words (in hex): 63 (which is 0x3F) is 0x20 | 0x10 | 0x8 | 0x4 | 0x2 | 0x1. If you look at them all in binary, it is obvious:

0x20 : 00100000
0x10 : 00010000
0x08 : 00001000
0x04 : 00000100
0x02 : 00000010
0x01 : 00000001

And 63 is:

0x3F : 00111111

---

If you're getting some return status and want to know what it means, you'll have to use binary and. For example:

if (status & 0x02) 
{
}

Will execute if the flag 0x02 (that is, the 2nd bit from the right) is turned on in the returned status. Most often, these flags have names (descriptions), so the code above will read something like:

if (status & CONNECT_ERROR_FLAG)
{
}

Again, the status can be a combination of stuff:

// Check if both flags are set in the status
if (status & (CONNECT_ERROR_FLAG | WRONG_IP_FLAG))
{
}

P.S.: To learn why this works, this is a nice article about binary flags and their combinations.

Answer 2

63 (decimal) equals 0x3F (hex). So 63 is a combination of all of the following flags:

0x20
0x10
0x08
0x04
0x02
0x01

Is that what you were looking for?

Answer 3

I'd give you the same answer as Chris: your return value 0x63 seems like a combination of all the flags you mention in your list (except 0x0).

When dealing with flags, one easy way to figure out by hand which flags are set is by converting all numbers to their binary representation. This is especially easy if you already have the numbers in hexadecimal, since every digit corresponds to four bits. First, your list of numbers:

   0x01           0x02           0x04        ...        0x20
    |              |              |                      |
    |              |              |                      |
    V              V              V                      V
0000 0001      0000 0010      0000 0100      ...     0010 0000

Now, if you take your value 63, which is 0x3F (= 3 * 16¹ + F * 16⁰, where F = 15) in hexadecimal, it becomes:

   0x3F
    |
    |
    V
0011 1111

You quickly see that the lower 6 bits are all set, which is an "additive" combination (bitwise OR) of the above binary numbers.