Proper way to access array members as a different type?

Question

I have a large array of uint16_t.

Most of its members are uint16_t, but some are int16_t and some uint8_t.

How would you handle that?

---

By the way, I tried:

1. Pointers:

   Used 2 pointers, one int16_t* and the other uint8_t*, both initialized to the start of the array, to access member of the array that are int16_t and uint8_t.

   (That worked initially, but I ran into problems when later in the program something else changed the value of the pointers, so I don't trust it.)

2. Type definition with a union.

   In file.h:

   typedef union {
     uint16_t  u16[NO_OF_WORDS];     // As uint16_t
     int16_t   s16[NO_OF_WORDS];     // As int16_t
     uint8_t   u8[2 * NO_OF_WORDS];  // As uint8_t
   } ram_params_t;
   extern ram_params_t ram_params[];
   

   In file.c:

   ram_params_t ram_params[] = {0};
   

   (That really bombed.)

3. Casting.

   (I didn't get very far with that.)

Answer 1

The problem with your attempt #2 was that you made an array of arrays.

Either do this:

typedef union {
  uint16_t  u16;     // As uint16_t
  int16_t   s16;     // As int16_t
  uint8_t   u8[2];   // As uint8_t
} ram_params_t;
extern ram_params_t ram_params[NO_OF_WORDS];

ram_params_t ram_params[NO_OF_WORDS];

uval16  = ram_params[i].u16;
sval16  = ram_params[i].s16;
uval8_1 = ram_params[i].u8[0];
uval8_2 = ram_params[i].u8[1];

Or you do this:

typedef union {
  uint16_t  u16[NO_OF_WORDS];     // As uint16_t
  int16_t   s16[NO_OF_WORDS];     // As int16_t
  uint8_t   u8[2 * NO_OF_WORDS];  // As uint8_t
} ram_params_t;
extern ram_params_t ram_params;

ram_params_t ram_params;

uval16  = ram_params.u16[i];
sval16  = ram_params.s16[i];
uval8_1 = ram_params.u8[2*i];
uval8_2 = ram_params.u8[2*i+1];

I don't see anything wrong with your attempt #1 either. I think that I would probably do that rather than using a union.

Answer 2

The array elements of other types must be the same size as the uint16_t so just cast them. In the case of the 8-bit data, there is a possibility that the upper 8 bits are undefined, so I masked them off.

#include <stdio.h>

#define uint16_t    unsigned short
#define int16_t     short
#define uint8_t     unsigned char

int main() {
    uint16_t n;

    // convert uint16_t to int16_t
    n = 0xFFFF;
    printf ("%d\n", (int16_t)n);

    // convert uint16_t to uint8_t
    n = 168;
    printf ("%c\n", (uint8_t)(n & 0xFF));

    return 0;
}

Program output is

-1
¿