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);
 } 

• Juraj: uint8_t* is a pointer. on AVR it is 2 bytes. on 32bit architectures 4 bytes
• Juraj: you want const uint8_t* const neighbourSet?
• Mikael Patel: Hint: sizeof() gives the number of bytes. You are looking for membersof(). It is often defined as #define membersof(x) (sizeof(x) / sizeof(x[0])).
• Anum Sheraz: int totalPeers = (sizeof(neighbourSet) / sizeof(neighbourSet[0])); worked. Thanks Mikael !
• RubberDuck: @MikaelPatel you should post that as an answer. I was going to post the same, but then I saw your comment.

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.

• Pararera: Sizeof doesn't work for pointers