How do I utilize an array of linked lists?

[sharrakor]

The concept is easy enough to get. I just can't find the paper I wrote down how to do it on...

So my question is just how to make an element of an array of linked lists refer to an element of the list.

So if I have:

struct list[20]
{
int num;
struct list *next;
}

Is that the proper way to start one off?

And how do I refer to an element inside? Something like head[2]->next, or head[2].next?

Thanks!

[Salem]

struct list {
    int num;
    struct list *next;
} arrayOfLists[50];

The rest is OK.

But use a typedef to separate the type declaration from the instance of a variable of that type.

[sharrakor]

Thanks for the help.

But I still don't know how to actually call an element of a particular list. Is it something like:

test[3]->num=5;

//or

test[3].num=5;

All I need is 1 line of code to see how this is done

[Salem]

Try it.
One compiles, the other doesn't.

But FWIW, an array of linked lists would be more like
struct list *arrayOfLists[20];

[MK27]

I would not use an array to deal with a linked list -- there is no point. Part of the idea of a linked list is that you do not use any particular fixed variable (or array) to refer to "an element" (so call it a node, not an element). You use temporary pointers, and by assigning memory with malloc. Normally, if you did this and then threw the pointer away or reassigned it, this memory would be considered "leaked". But with a linked list, you can find a chunk of memory by using the pointers included in the node structure. So the answer to your last question is you always use -> because *node is always a pointer. Look:

#include <stdio.h>
#include <stdlib.h>

typedef struct _node {
	int X;
	struct _node *next;
} node;

void addnode (node *head, int value) {
	node *new=malloc(sizeof(node)), *ptr=head;
	new->X=value; new->next=NULL;
	while (ptr->next) ptr=ptr->next;  /* find the "tail" */
	ptr->next=new;
}

void showlist (node *head) {
	node *ptr=head;
	while (ptr->next) {
		printf ("%d\n",ptr->X);
		ptr=ptr->next;
	}
	printf ("%d\n", ptr->X);
}
	

int main() {
	node *head=malloc(sizeof(node));
	int i;
	head->X=100;
	for (i=0; i<5; i++) addnode(head,(i+2)*100);
	showlist(head);	
	return 0;
}

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That's a singly linked list because there is no "previous" pointer. They are great way to improve your understanding of pointers and memory assignment. Notice how head is used; this is the essence of the linked list method.

[sharrakor]

Thanks a lot both of you, especially MK. That's a nice, simple example of a linked list.

We have to use them for my programming class, so I don't have much of a choice. It wouldn't surprise me if it was just to improve understanding of pointers and memory assignment.

By the way, head[2]->next; and head[2].next aren't working. I tried them before I posted.

[MK27]

By the way, head[2]->next; and head[2].next aren't working. I tried them before I posted.

In what context?

Again, if you are using an array to deal with a linked list, I would say you are barking up the wrong tree. There are important and significant differences between between an array and a linked list. They are not the same thing.

So you should not have a head[2]. Or a head[0], etc. The "head" is the first node in the list, NOT the first element of an array.

[sharrakor]

Got it. Thanks for the time guys.