What does this line of code do? Const uint32_t goodguys = 0x1 << 0 - tags

Question

Can someone tell me what is being done here:
Const uint32_t goodguys = 0x1 << 0
I'm assuming it is c++ and it is assigning a tag to a group but I have never seen this done. I am a self taught objective c guy and this just looks very foreign to me.

Answer 1

Well, if there are more lines that look like this that follow the one that you posted, then they could be bitmasks.
For example, if you have the following:
const uint32_t bit_0 = 0x1 << 0;
const uint32_t bit_1 = 0x1 << 1;
const uint32_t bit_2 = 0x1 << 2;
...
then you could use use the bitwise & operator with bit_0, bit_1, bit_2, ... and another number in order to see which bits in that other number are turned on.
const uint32_t num = 5;
...
bool bit_0_on = (num & bit_0) != 0;
bool bit_1_on = (num & bit_1) != 0;
bool bit_2_on = (num & bit_2) != 0;
...
So your 0x1 is simply a way to designate that goodguys is a bitmask, because the hexadecimal 0x designator shows that the author of the code is thinking specifically about bits, instead of decimal digits. And then the << 0 is used to change exactly what the bitmask is masking (you just change the 0 to a 1, 2, etc.).

Answer 2

Although base 10 is a normal way to write numbers in a program, sometimes you want to express the number in octal base or hex base. To write numbers in octal, precede the value with a 0. Thus, 023, really means 19 in base 10. To write numbers in hex, precede the value with a 0x or 0X. Thus, 0x23, really means 35 in base 10.
So
goodguys = 0x1;
really means the same as
goodguys = 1;
The bitwise shift operators shift their first operand left (<<) or right (>>) by the number of positions the second operand specifies. Look at the following two statements
goodguys = 0x1;
goodguys << 2;
The first statement is the same as goodguys = 1;
The second statement says that we should shift the bits to the left by 2 positions. So we end up with
goodguys = 0x100
which is the same as goodguys = 4;
Now you can express the two statements
goodguys = 0x1;
goodguys << 2;
as a single statement
goodguys = 0x1 << 2;
which is similar to what you have. But if you are unfamiliar with hex notation and bitwise shift operators it will look intimidating.
When const is used with a variable, it uses the following syntax:
const variable-name = value;
In this case, the const modifier allows you to assign an initial value to a variable that cannot later be changed by the program. For Instance
const int POWER_UPS = 4;
will assign 4 to variable POWER_UPS. But if you later try to overwrite this value like
POWER_UPS = 8;
you will get a compilation error.
Finally the uint32_t means 32-bit unsigned int type. You will use it when you want to make sure that your variable is 32 bits long and nothing else.