Initializing Structs

Struct with flexible array members (FAM)

Here’s a struct which represents a “list”.

typedef struct List_ {
   size_t len;
   short elmts[];
} List;

Note that elmts is an array. How many elements does it take? Until it is not initialized to set aside memory so it can hold \$n\$ elements, that question cannot be answered.

How does one initialize it (allocate memory) so that it can take \$n\$ elements? Use malloc (stdlib.h) and compute the size of List plus the size of the type of elmts times how many elements we want elmts to be able to store.

Let’s exemplify with a function that creates and initializes a new list and then returns it:

List *list_new(size_t len, short elmts[]) {
  /* Allocate memory for the list, accounting for the space
   * necessary for the `len` array elements of type `short`. */
  List *list = malloc(sizeof(List) + sizeof(short) * len);

  /* Set the length. */
  list->length = len;

  /* Copy the elements one by one. */
  for (size_t i = 0; i < len; ++i)
    list->elmts[i] = elmts[i];

  return list;

Then we can use it like this:

List *list = list_new(3, (short []){10, 20, 30})

for (size_t i = 0; i < 3, ++i)
  printf("%d\n", list->elmts[i]);


As always, when using dynamic memory allocation, one must remember to free() the memory when it is no longer needed.