C: setting up linked list using malloc()

I'm completely new to linked lists and have been give the below task:

Write a program that creates a Linked List of 10 int values (ranging from 1 to 10) using malloc().

" (There are further parts to this question but don't relate to what I'm stuck on)

I understand the concept of a linked list and I know how to use malloc(). However, the part I don't understand is nodes.

For example: how do you use malloc() on the nodes and the meaning of ->. If someone could explain how the nodes work to set up the linked list of 10 int values that would be great.

Solution

What you basically need to understand is that you need to create several nodes. Let's think about 4 nodes. Node A, B, C and D. So the use of malloc()is going to be to save in memory this 4 nodes (not in order, obviously).

It is a linked list, not a sequential list. This means that you should be able to "access" it like you would normally do in a sequential list, but they are not sequentially saved in memory (hardware speaking).

So: Your node A will have a pointer. This pointer will point to Node B. Same goes to Node B to node C, and Node C to Node D.

A->B->C->D

Remember ALL NODES must have whatever content you want, and a pointer to the next node, so you can access it from there.

This means that Node A can be in (imagine), position 4 in memory, Node B in position 16 in memory, Node C in position 2 in memory, and node D in position 300 in memory.

For example, an easy example using struct:

struct Node {
   int data;
   struct Node *next;
};

When you insert a new node, you just need to change the pointers:

A->B->C->D E (you want to insert it in the second position).

So you need to change pointers. Something like:

 E=A->nextNode; //A is no longer pointing to B
B=E->nextNode; //Now E is pointing to B

To use malloc as you are asking:

struct Node* A= malloc(sizeof(struct Node));   
  struct Node* B = malloc(sizeof(struct Node));

Take a look in here to see it better