How can I declare and initialize an array of pointers to a structure in C?

Question

I have a small assignment in C. I am trying to create an array of pointers to a structure. My question is how can I initialize each pointer to NULL? Also, after I allocate memory for a member of the array, I can not assign values to the structure to which the array element points.

#include <stdio.h>
#include <stdlib.h>

typedef struct list_node list_node_t;

struct list_node
{
   char *key;
   int value;
   list_node_t *next;
};


int main()
{

   list_node_t *ptr = (list_node_t*) malloc(sizeof(list_node_t));

   ptr->key = "Hello There";
   ptr->value = 1;
   ptr->next = NULL;

   // Above works fine

   // Below is erroneous 

   list_node_t **array[10] = {NULL};      

   *array[0] =  (list_node_t*) malloc(sizeof(list_node_t));
    array[0]->key = "Hello world!";  //request for member ‘key’ in something not a structure or union
    array[0]->value = 22;            //request for member ‘value’ in something not a structure or union 
    array[0]->next = NULL;           //request for member ‘next’ in something not a structure or union


    // Do something with the data at hand
    // Deallocate memory using function free 

   return 0;
}

Answer 1

Here:

list_node_t **array[10] = {NULL};

You're declaring an array of 10 pointers to pointers to your struct. What you want is an array of 10 pointers to your struct:

list_node_t *array[10] = {NULL};

It's confusing because yes, array really is a pointer to a pointer, but the square bracket notation sort of abstracts that away for you in C, and so you should think of array as just an array of pointers.

You also don't need to use the dereference operator on this line:

*array[0] =  (list_node_t*) malloc(sizeof(list_node_t));

Because C dereferences it for you with its bracket notation. So it should be:

array[0] =  (list_node_t*) malloc(sizeof(list_node_t));