Question

I am a linguist in charge of a C program, so please excuse me if the answer is obvious. I have the following code:

typedef struct array_s {
  (...)
  void **value;
} array_t;
typedef array_t *array_pt;

array_pt array_new (int size) {
  (...)
  array->value = (void **)malloc(size*sizeof(void *));
}

void* array_get (array_pt arr, int i) {
  return arr->value[i];
}

int main () {
  int a = 1234;
  int *ptr = &a;

  array_pt array = array_new(1);
  array_add(array, ptr);

  printf("%i\n", (int)array_get(array, 0));
}

It is supposed to provide me with a multi-purpose array (for storing int and char*, if I understood I can only use void), and I guess there are no problems of allocating/freeing. However, I cannot get to cast it into anything useful (i.e., get back the “original” int/char*), and for what I understood it could be because I am in a 64-bit system and the size of a pointer to void is different from the size of a pointer to int/char* (the program is supposed to be used in both 64 and 32 bit systems). I tried using intptr_t and other alternatives, to no luck.

How can I be sure that the code will accept any data type and work on both 32 and 64 bit systems? Thank you.

EDIT:

Sorry for not adding array_add, here it is:

unsigned int array_add (array_pt array, void *ptr) {
  (...) // get the next index
  // allocate if needed
  array->value = (void **)realloc(array->value, array->size*sizeof(void *));

  array->value[index] = p;
}

Answer 1

You need to dereference your pointer:

int* temp = array_get(array, 0);
printf("%i\n", *temp);

However, I strongly recommend avoiding this type of approach. You’re basically giving away the small amount of help the compiler in C will normally provide – purposefully trying to make non-typesafe arrays.