Using malloc in C to allocate space for a typedef'd type

Question

I\'m not sure exactly what I need to use as an argument to malloc to allocate space in the table_allocate(int) function. I was thinking just

Answer 1

count_table* cTable = malloc(sizeof(count_table*))

is wrong. It should be

count_table* cTable = malloc(sizeof(count_table));

Also, you must allocate memory for list_node_t also seperately.

EDIT:

Apart from what Clifford has pointed about allocating memory for the list node, I think the memory allocation should also be taken care for the char *key inside of the list node.

Answer 2

In addition to the other posters who point out that you're only allocating enough space for the pointer, not the space the data you want will occupy, I strongly urge you to do things like this:

count_table* cTable = malloc(sizeof(*cTable));

This will help you in case the type of cTable ever changes, you won't have to adjust two parts to that line, just the type.

Answer 3

Given that the int is a "size" parameter for the created count_table_t, it appears that you are supposed to both allocate the count_table_t itself, as well as initialise its members.

Initialising the list_array member also involves a memory allocation, so it would look like:

count_table_t *table_allocate(int size)
{
    count_table_t *table = malloc(sizeof *table);
    int i;

    table->size = size;
    table->list_array = malloc(size * sizeof table->list_array[0]);
    for (i = 0; i < size; i++)
        table->list_array[i] = NULL;

    return table;
}

However, you also need to check for some error conditions: the multiplication of size by sizeof table->list_array[0] could overflow, and either of the malloc() calls could fail. So the function should actually look like this:

count_table_t *table_allocate(int size)
{
    count_table_t *table;
    int i;

    /* Check for overflow in list allocation size */
    if (size < 0 || size > (size_t)-1 / sizeof table->list_array[0])
        return NULL;

    table = malloc(sizeof *table);

    if (table == NULL)
        return NULL;

    table->size = size;
    table->list_array = malloc(size * sizeof table->list_array[0]);

    if (table->list_array == NULL) {
        free(table);
        return NULL;
    }

    for (i = 0; i < size; i++)
        table->list_array[i] = NULL;

    return table;
}

(Note that (size_t)-1 is a constant equal to the maximum value of a size_t, which is the type of the parameter to malloc()).

Answer 4

Your suggestion: count_table* cTable = malloc(sizeof(count_table*)) would only allocate space for a pointer to a count_table.

You'd need

count_table* cTable = malloc(sizeof(count_table) ) ;

Each list node would be separately allocated and cTable->size and cTable->list_array and the last list_node_t::next updated accordingly. Maintaining a pointer to the last node added would make adding nodes faster.

I am not sure why count_table::list_array is of type list_node_t** rather than just list_node_t* (and equally called list_array rather than just list). Is it your intention that it is both an array and a list at the same time? That would be somewhat redundant. The member need only be a pointer to the first node, successive nodes are then accessed via list_node::next