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.

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*.

Quote:

---

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.

Quote:

---

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.

Quote:

---

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?

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.

---

The first part has been mentioned by tabstop. For the malloc thing, I think earlier it would be returning pointer to something else, that's why they typecast it. BTW it's just a wild guess.

Quote:

---

Originally Posted by bengreenwood

So why do you think they bothered casting it in the examples?

---

Historically, there was no void pointer and the cast was necessary:

Cprogramming.com FAQ > Casting malloc

however this is no longer the case.

Thanks for the advice all. Makes sense.