Question

I want to ‘construct’ (read: malloc and memset) my hashtable in c. To do this, I created a function as follows:

int maketable(struct hash_entry **table, int size){

    table = (struct hash_entry **)malloc(size*sizeof(struct hash_entry *));
    int i = 0;
    for (; i<size; i++) {
        memset(table[i], '\0', sizeof(struct hash_entry *));
    }
    return 0;

}

Given that table will have been declared as

struct hash_entry **table[size]

before entering this code, I won’t lose anything when I return from maketable, right?

EDIT:
Does passing table into maketable() ensure that as long as I change the data that table points to, the changes will be preserved?

EDIT II:
I am trying to allocate an array of pointers to pointers to hash_entries

Answer 1

Your code is assigning to the local table variable, the caller is not affected. This results in a memory leak.

Outside the function, you have declared table as an array of pointer to pointers of struct hash_entry – I’m guessing you rather just want an array of pointers to struct hash entries.

If you actually declare table as an array, there’s no need to malloc that space. You just need a loop to set every element in it to NULL (Don’t memset each element to zeroes).

If the goal is to allocate the entire table, this is perhps what you’re looking for:

struct hash_entry **table;
...
int maketable(struct hash_entry ***table, int size){

    *table = malloc(size* sizeof **table);
    int i = 0;
    for (; i<size; i++) {
       (*table)[i] = NULL;
    }
    return 0;
}

call it like

maketable(&table,100);

I’d rather make that return the table like so:

struct hash_entry ** maketable(int size){
   return calloc(size, sizeof(struct hash_entry *));
}

if the declaration struct hash_entry **table[size] is really what you want, you need to tell us what your maketable() function actually is supposed to do (e.g. do you want to a dynamically allocated ‘array’ as one of the elements in that table ?