How to initialise a pointer to pointer struct in C?

Question

I have a struct which is a node, and another which is a list of these nodes. In the list struct, its an array of nodes, but instead of an array, it's a pointer to pointer with a size integer:

typedef struct node {
    struct node *next;
    MyDef *entry;
} Node;


typedef struct list {
    Node **table;
    int size;
} List;

List *initialize(void)
{
    List *l;
    Node **n;

    if ((l = (List *)malloc(sizeof(List))) == NULL)
        return NULL;
    l->size = 11;

    /* I think this is correctly allocating the memory for this 'array' of nodes */
    if ((n = (Node **)malloc(l->size * sizeof(Node))) == NULL)
        return NULL;

    /* Now, how do I set MyDef *entry and Node *next to NULL for each of the 'array'? */

    l->table = n;

    return l;
}

How do I set MyDef *entry and Node *next to NULL for each of the 'array'?

Answer 1

First of all, it seems to me that you have an error in your code allocating the array: It should say sizeof(Node*) rather than sizeof(Node) as you want to allocate an array of pointers to Node not an array of Node objects.

Then you can iterate through the array list:

for ( unsigned i = 0; i < l->size; ++i )
{
    Node* node = l->table[ i ];
    node->entry = NULL;
    node->next = NULL;
}

Another hint: You really should check your initialization function against possibilities of memory leaks.

Answer 2

(Node **) is pointer to [array of] pointer to Node, so array you allocate will not have any struct members.

You should use (Node *) and then you'll have pointed array of Node structs, or allocate each Node separately, then place pointers to them into your array.

There's exist function calloc() in standard C library for your case: it inits allocated area with 0's (which corresponds to (char/short/int/long)0, 0.0 and NULL).

Also there's a memory leak.

/* I think this is correctly allocating the memory for this 'array' of nodes */
if (... == NULL)
    return NULL;

When array allocation fails you do not free List, but lose pointer to it. Rewrite it as:

/* I think this is correctly allocating the memory for this 'array' of nodes */
if ((n = (Node **)malloc(l->size * sizeof(Node))) == NULL) {
    free(l);
    return NULL;
}

So from my point of wiev correct code would be:

typedef struct node {
    struct node *next;
    MyDef *entry;
} Node;


typedef struct list {
    Node *table; /* (!) single asterisk */
    int size;
} List;

List *initialize(void)
{
    List *l;
    Node **n;

    if ((l = (MList *)malloc(sizeof(List))) == NULL)
        return NULL;
    l->size = 11;

    /* I think this is correctly allocating the memory for this 'array' of nodes */
    if ((n = (Node *)calloc(l->size, sizeof(Node))) == NULL)
    {
        free(l);
        return NULL;
    }

    /* Now, how do I set MyDef *entry and Node *next to NULL for each of the 'array'? */

    l->table = n;

    return l;
}

Futhermore C99 allows you to make variable size structs, so you able to init struct like

typedef struct list {
    int size;
    Node table[0]
} List;

And allocate as many Node's in table as you want using malloc(sizeof(List) + sizeof(Node)*n);