Thread: Question About Linked Lists Example

Hi,

I've been working my way through some examples about how to program C, and I think I'm beginning to understand the basics of linked lists. In the following tutorial, they use some temporary variables to work their way through a linked list.

C7

However, the full code isn't shown, and the link to the example code doesn't work. So I have a question.

Before this code, they must have declared and allocated memory for "current_node" first, right?

Code:

// First deal with the base case, setting the head pointer
head = (struct list_node *) malloc(sizeof(struct list_node));
head->node_id = 1;

// Set a temp variable to our current last node
current_node = head;

When I made a program based on the example, I did the following first:

Code:

struct person *current_node;
current_node = (struct person *) malloc(sizeof(struct person)); // allocate memory

Was I right to do this? It seems like I had to. Similarly, at the bottom of the tutorial, they do this:

Code:

// Copy the head pointer
temp = head;

Again, you would need to do something like this first, right?

Code:

struct person *temp;
temp = (struct person *) malloc(sizeof(struct person)); // allocate memory

Thanks.

Originally Posted by BEN10

You only do malloc when you want a new node to be added to the list. Here

Code:

current_node = head;

You are only assigning the address where head points to current_node, so no need to make a room for it. It's only pointing to a node for which room has been initially allocated.
One more thing while doing malloc, no need to cast it, it itself returns void*.

Don't you mean where current_node points to head? Was that a typo, or am I getting mixed up somewhere?

Also with the malloc thing, I just based my code on the examples. I tried what you said, and it worked. So why do you think they bothered casting it in the examples?

Thanks for the advice.

Originally Posted by bengreenwood

Don't you mean where current_node points to head? Was that a typo, or am I getting mixed up somewhere?

Assignments move from right to left, so the value on the right (head) is assigned to the value on the leftg (current_head), so the address where head points is assigned to current_node.

Thanks for the advice all. Makes sense.