创建链接列表时是否不需要创建实际节点？

Question

I'm studying linked lists from this article.

The writer of the tutorial never creates actual nodes, but only pointer variables of type node, as you can see with the following code...

struct node* head = NULL;
struct node* second = NULL;
struct node* third = NULL;

and then he allocates space for them in the heap...

head = (struct node*)malloc(sizeof(struct node)); 
second = (struct node*)malloc(sizeof(struct node));
third = (struct node*)malloc(sizeof(struct node));

why doesn't he ever create actual nodes? code for which should look something like this...

struct node head;
struct node second;
struct node third;

If my knowledge is right (correct me if I'm wrong). Simply declaring the pointer variable doesn't create the actual variable(node in case of linked lists), and hence cannot be dereferenced like the writer of the tutorial in the article using the code

head->data = 1;

I mean, if that works, then why doesn't this work?

int *a;
a=5;
printf("%d",*a);

Obviously, the above code doesn't output 5.

Which means that another variable needs to be created, and then it needs to be stated that the address of the variable is stored in a pointer variable, only then it can be dereferenced...like the following code...

int *a;
int b=5;
a=&b;
printf("%d",*a);

This outputs 5.

So how does the author gets away with not creating the nodes? He simply creates the pointer variables and then simply dereferences them....

Answer 1

The nodes are in the heap, that's what malloc is for.

To use code without linked list to explain, it's similar to:

int *a = NULL;
a = malloc(sizeof(int));
*a = 5;
printf("%d",*a);

Answer 2

The structure pointers are created so that they can refer to address.Since normal structure variables can't store the address, this syntax is wrong. struct node head; struct node second; struct node third;