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Home/ Questions/Q 3301960
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T20:48:13+00:00 2026-05-17T20:48:13+00:00

I have a struct which is a node, and another which is a list

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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’?

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  1. Editorial Team
    Editorial Team
    2026-05-17T20:48:13+00:00Added an answer on May 17, 2026 at 8:48 pm

    (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);

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