Question about "fclone_ref = (atomic_t *) (skb + 2)" in kfree_skbmem

["Huangqiang Zhou" <linuxkernel.xqiang@xxxxxxxxx>]

Hi all:
	My questions are as follows.

///////////////////< code from linux-2.6.35.4
/*
 *	Free an skbuff by memory without cleaning the state.
 */
static void kfree_skbmem(struct sk_buff *skb)
{
	struct sk_buff *other;
	atomic_t *fclone_ref;

	switch (skb->fclone) {
	case SKB_FCLONE_UNAVAILABLE:
		kmem_cache_free(skbuff_head_cache, skb);
		break;

	case SKB_FCLONE_ORIG:
		fclone_ref = (atomic_t *) (skb + 2);      ///////////////////////////< question here
		if (atomic_dec_and_test(fclone_ref))
			kmem_cache_free(skbuff_fclone_cache, skb);
		break;

	case SKB_FCLONE_CLONE:
		fclone_ref = (atomic_t *) (skb + 1);    /////////////////////////////< question here
		other = skb - 1;

		/* The clone portion is available for
		 * fast-cloning again.
		 */
		skb->fclone = SKB_FCLONE_UNAVAILABLE;

		if (atomic_dec_and_test(fclone_ref))
			kmem_cache_free(skbuff_fclone_cache, other);
		break;
	}
}

	skb->|--------------------|
	     |struct sk_buff *next| 
	     |--------------------|
	     |struct sk_buff *prev| 
	     |--------------------|
	     |      ......        | 
	     |--------------------|
	     |  automic_t users   | 
  skb+1->|--------------------|	fclone_ref = (atomic_t *) (skb + 1);
	     |struct sk_buff *next| 
	     |--------------------|
	     |struct sk_buff *prve| 
	     |--------------------|
	     |      ......        | 
	     |--------------------|
	     |  automic_t users   | 
  skb+2->|--------------------|	fclone_ref = (atomic_t *) (skb + 2);	
	     |struct sk_buff *next| 
	     |--------------------|
	     |struct sk_buff *prve| 
	     |--------------------|
	     |      ......        | 
	     |--------------------|
	     |  automic_t users   | 
	     |--------------------|

typedef struct {
	int counter;
} atomic_t;

so when skb+1, it poins to  the member next. if we use  "(atomic_t *) (skb + 1);", what the hell result is? and how can we visit the member "counter" of atomic_t?
			
Thanks
 
Huangqiang Zhou
------------------	
2010-09-19


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