How to initialize structure having unint_t as flexible array member?

Question

I define a struct, have a one-member the type is uint8_t, this member store mac address.

Programming on Arduino IDE.

Struct:

typedef struct devInfo{
  uint8_t address[];
  unsigned int count;
  unsigned int filePos;
}struct_devInfo;

• Quesion 1:

I use these two ways to give a value but cannot assign value to variable.

Method 1 > struct_devInfo slave = {{0xA1, 0xB2, 0xC3, 0xD4, 0xE5, 0xF6}, 0, 0};

Method 2 > slave.address[] = {0xA1, 0xB2, 0xC3, 0xD4, 0xE5, 0xF6};

How can I access this type of value?

• Question 2:

I will use the variables of this structure, salve1, slave2... etc.

In addition to the structure, is there some better way? Can you demonstrate?

Answer 1

A struct with a flexible array member needs to be the last member in the struct. Change to

typedef struct devInfo{
  unsigned int count;
  unsigned int filePos;
  uint8_t address[];
}struct_devInfo;

Then you need to allocate enough memory:

struct_devInfo *p = malloc(sizeof *p + size);

Then you could do this:

const uint8_t initArr[] = {0xA1, 0xB2, 0xC3, 0xD4, 0xE5, 0xF6};
memcpy(p, initArr, sizeof initArr);

But since it seems to be a field that does not require a flexible member, I'd use this instead:

typedef struct devInfo{
  unsigned int count;
  unsigned int filePos;
  uint8_t address[6]; // Specify the size
}struct_devInfo;

Then you don't need to allocate memory