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

Question

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.

AmiguelS · Accepted Answer

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

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

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