Length of uint8_t* const array

Question

I've an array of type uint8_t* const with 6 items, defined like this

uint8_t* const neighbourSet[] = {PEER1, PEER2, PEER3, PEER4, PEER5, PEER6};

Whereas each element in this array is static uint8_t defined like this

static uint8_t PEER1[] {0x86, 0xF3, 0xEB, 0x7A, 0xE8, 0x3B};
static uint8_t PEER2[] {0x86, 0xF3, 0xEB, 0x7A, 0xA1, 0x09};
static uint8_t PEER3[] {0x84, 0x0D, 0x8E, 0x03, 0x95, 0xED};
static uint8_t PEER4[] {0x84, 0x0D, 0x8E, 0x03, 0x99, 0xD5};
static uint8_t PEER5[] {0x80, 0x7D, 0x3A, 0xC5, 0x2B, 0x79};
static uint8_t PEER6[] {0x84, 0x0D, 0x8E, 0x03, 0x95, 0x1D};

When I do

  Serial.print("size:");
  Serial.println(sizeof(neighbourSet));

I get size:24. I am expecting the size to be 6. could someone explain ? I don't want to hardcode the total number of PEER in a separate integer variable. I want my for loop to run through each element till the total size of array.

    for (int i = 0; i < sizeof(neighbourSet); i++) {
         //do something for each PEER. 
         delay(50);
    }   

Answer 1

sizeof() gives the number of bytes. You are looking for membersof(). It is often defined as:

#define membersof(x) (sizeof(x) / sizeof(x[0]))

And in the example above:

for (int i = 0; i < membersof(neighbourSet); i++) {
     //do something for each PEER. 
     delay(50);
}  

For more practical C/C++ macro and type extensions please see the Cosa Types.h file.