Why does this particular flag involve an "or" operation?

[Mitch996]

uint8_t flags = BLE_GAP_ADV_FLAG_BR_EDR_NOT_SUPPORTED |
                BLE_GAP_ADV_FLAGS_LE_ONLY_GENERAL_DISC_MODE;

And then when I look up "BLE_GAP_ADV_FLAGS_LE_ONLY_GENERAL_DISC_MODE;", I got this:

(BLE_GAP_ADV_FLAG_LE_GENERAL_DISC_MODE | BLE_GAP_ADV_FLAG_BR_EDR_NOT_SUPPORTED)   /**< LE General Discoverable Mode, BR/EDR not supported. */

I'm having a bit of brain stall right now and can't figure it out. I know it's a trick to make the whole thing a "switch" of some sort but really don't know how.

[wlgrd]

I don't really understand the question, and why you set your flags like this. If using the LE_ONLY flag you implicitly also set BLE_GAP_ADV_FLAG_BR_EDR_NOT_SUPPORTED.

#define BLE_GAP_ADV_FLAG_LE_GENERAL_DISC_MODE (0x02) Note that this define does not say LE_ONLY.

#define BLE_GAP_ADV_FLAGS_LE_ONLY_GENERAL_DISC_MODE (BLE_GAP_ADV_FLAG_LE_GENERAL_DISC_MODE | BLE_GAP_ADV_FLAG_BR_EDR_NOT_SUPPORTED)

Hence, if you want to set your flag like this, either change it to

uint8_t flags = BLE_GAP_ADV_FLAGS_LE_ONLY_GENERAL_DISC_MODE

or

uint8_t flags = BLE_GAP_ADV_FLAG_LE_GENERAL_DISC_MODE | BLE_GAP_ADV_FLAG_BR_EDR_NOT_SUPPORTED)

[Mitch996]

I forgot to say that I was reading someone else's code. If it were me I wouldn't use the "|" operator like that. My question isn't with the protocol itself, but on the "|" operator, I've seen this use several times before but I didn't really care why or how until now.

[AmiguelS]

The OR operator is being used to set multiple bits of that flag in a single line.

For example, conside you have a flag of 1 byte (8 bits), and each bit controls one feature.

#define FEATURE_ONE   0b00000001
#define FEATURE_TWO   0b00000010
#define FEATURE_THREE 0b00000100
#define FEATURE_FOUR  0b00001000
#define FEATURE_FIVE  0b00010000
#define FEATURE_SIX   0b00100000
#define FEATURE_SEVEN 0b01000000
#define FEATURE_EIGHT 0b10000000

And now you wish to turn on features 2 and 5. You can do so with

uint8_t flags = FEATURE_TWO | FEATURE_FIVE;

This works because when doing an OR of the two numbers, the set bits from each number prevail, so:

    00000010
    00010000
or__________
    00010010

[Mitch996]

Got it. Much obliged.

[ovrebekk]

Please accept one of the answers, to show the case as closed.