Why the first node of a linked list is declared as a pointer?

Now I know that why pointers are used in defining linked lists. Simply because structure cannot have a recursive definition and if there would have been no pointers, the compiler won't be able to calculate the size of the node structure.

struct list{
        int data;
        struct list* next;    // this is fine
};

But confusion creeps up when I declare the first node of the linked list as:

struct list* head;

Why this has to be a pointer? Can't it be simply declared as

struct list head;

and the address of this used for further uses? Please clarify my doubt.

Solution

There's no definitive answer to this question. You can do it either way. The answer to this question depends on how you want to organize your linked list and how you want to represent an empty list.

You have two choices:

A list without a "dummy" head element. In this case the empty list is represented by null in head pointer
```
struct list* head = NULL;
```
So this is the answer to your question: we declare it as a pointer to be able to represent an empty list by setting head pointer to null.
A list with a "dummy" head element. In this case the first element of the list is not used to store actual user data: it simply serves as a starting "dummy" element of the list. It is declared as
```
struct list head = { 0 };
```
The above represents an empty list, since head.next is null and head object itself "does not count".

I.e. you can declare it that way, if you so desire. Just keep in mind that head is not really a list element. The actual elements begin after head.

And, as always, keep in mind that when you use non-dynamically-allocated objects, the lifetime of those objects is governed by scoping rules. If you want to override these rules and control the objects' lifetimes manually, then you have no other choice but to allocate them dynamically and, therefore, use pointers.