Cast void* to int* problem. I think my understanding of casting is flawed.

I've got my "write_vector" function finished, I think it works:

Code:

void *write_vector(struct vector *v, int index, void *data)
{
    struct data_node *cur_node;
    cur_node = v->data;
    if(v->elem > index)
    {
        cur_node = v->data;
        for(;index;index--)
            cur_node = cur_node->next;
    }
    else
    {
        int c;
        if(v->elem == 0)
            v->data = cur_node = (struct data_node*)malloc(sizeof(struct data_node));
        cur_node->data = calloc(1, v->size);
        for(c = 0;c < index;c++)
        {
            if(c >= v->elem && c < index)
            {
                cur_node->next = (struct data_node*)malloc(sizeof(struct data_node));
                cur_node->next->data = calloc(1, v->size);
            }
            cur_node = cur_node->next;
        }
        v->elem = index + 1;
    }
    cur_node->data = calloc(1, v->size);
    return memcpy(cur_node->data, data, v->size);
}

A pointer to the vector I'm writing to is passed in v, the index of the node to write to in index, and a pointer to the data to be copied to another location, which is stored in a node, is passed in data. It returns a pointer to the data in it's stored location.

And an int main just to debug by playing around with some vectors:

Code:

    int i = 5;
    struct vector *v = init_vector(sizeof(int));
    printf("%i\n", write_vector(v, 1, &i));
    printf("%i\n", v->elem);
    if(read_vector(v, 0) == NULL)
        printf("%s\n", getlasterrstr());
    printf("%i\n", (int*)(read_vector(v, 1)));

I think it's a problem with my understanding of casting, both line 3 and line 7 of main print the same value. The value changes every time I run the prog, so I'm going to guess that what I'm printing is the location of 5. How do I print the 5? Thanks for reading this far, a lot of code, I know .

My structs being used:

Code:

struct data_node{
    void* data;
    struct data_node *next;
};

struct vector{
    int elem;
    int size;
    struct data_node *data;
};

Okay, apparently I don't fully understand pointers when used in casting. Here's what main should look like: (minus some other debigging stuff)

Code:

int main(int argc, char **argv)
{
	int i = 5;
	struct vector *v = init_vector(sizeof(int));
	printf("%i\n", *((int*)write_vector(v, 1, &i)));
	printf("%i\n", *((int*)read_vector(v, 1)));

write_vector writes the 5 into memory(and also returns a pointer to it) and the read_vector returns a pointer to it.

Is there a standard vector lib for C?

It is clearer to assign to temporary pointer.

Code:

  int *p = write_vector(v,1,&i); 
 printf("%i\n", *p);      
 printf("%i\n",   *(int*)write_vector(v,1,&i)  );   // cast void pointer to int pointer and 
  // deference

Yeah, that looks a lot cleaner. Thanks for the pointer (hehe, I made a pun). I was getting a seg fault because it would fail to allocate nodes if the didn't exist, but I fixed it. Here's the completely working version:

Code:

void *write_vector(struct vector *v, int index, void *data)
{
	struct data_node *cur_node;
	cur_node = v->data; /* Set the current node to the root node */
	if(v->elem > index) /* Check to see if the index is within the bounds of the vector, if it is, do this the easy way */
	{
		cur_node = v->data;
		for(;index;index--)
			cur_node = cur_node->next;
	}
	else /* If it's not, do it the hard way */
	{
		int c;
		if(v->elem == 0) /* Check to see the root node exists */
		{
			v->data = cur_node = (struct data_node*)malloc(sizeof(struct data_node));
			cur_node->data = calloc(1, v->size);
		}
		for(c = 0;c < index;c++) /* Skip to index */
		{
			if((c + 1) >= v->elem && (c + 1) <= index) /* Does the next node exist, if not... */
			{
				cur_node->next = (struct data_node*)malloc(sizeof(struct data_node)); /* Allocate memory for it */
				cur_node->next->data = calloc(1, v->size); /* Allocate memory for it's data */
			}
			cur_node = cur_node->next;
		}
		if(index >= v->elem)
			v->elem = index + 1;
	}
	return memcpy(cur_node->data, data, v->size);
}