Question

I generally have ignored using macros while writing in C but I think I know fundamentals about them. While I was reading the source code of list in linux kernel, I saw something like that:

#define LIST_HEAD_INIT(name) { &(name), &(name) }
#define LIST_HEAD(name) \
    struct list_head name = LIST_HEAD_INIT(name)

(You can access the remaining part of the code from here.)

I didn’t understand the function of ampersands(I don’t think they are the address of operands here) in LIST_HEAD_INIT and so the use of LIST_HEAD_INIT in the code. I’d appreciate if someone can enlighten me.

Answer 1

To know what actually is happening we need the definition of struct list_head:

struct list_head {
        struct list_head *next, *prev;
};

Now consider the macros:

#define LIST_HEAD_INIT(name) { &(name), &(name) }
#define LIST_HEAD(name) struct list_head name = LIST_HEAD_INIT(name)

If in the code I write LIST_HEAD(foo) it gets expanded to:

struct list_head foo = { &(foo) , &(foo)}

which represents an empty doubly linked list with a header node where the next and prev pointers point to the header node itself.

It is same as doing:

struct list_head foo;
foo.next = &foo;
foo.prev = &foo;

So effectively these macros provide a way to initialize a doubly linked list.

And yes, & is used here as the address of operator.

EDIT:

Here is a working example

In the link provided by you. You had:

struct list_head test = LIST_HEAD (check);

which is incorrect. You should have:

LIST_HEAD (check);