Re: [NET] Possible bug in netif_receive_skb

[Serge Kuznetsov]

> 
> ->func() must either take or free up the SKB, there must be no
> violations of this rule.
> 

Could you explain it more clearly?

How it applies to that two ( even three ) scenarios, I've told?

What if we have the first scenario:

ptype_all->func = func1;
ptype_all->next = NULL;

  Will this function be called or not?

Second scenario:

ptype_all->func = func1;
ptype_all->next = &ptype1;

ptype1->func = func2;
ptype1->next = NULL;

Will func2() be called?

Third scenario:

ptype_all->func = func1;
ptype_all->next = &ptype1;

ptype1->func = func2;
ptype1->next = &ptype2;

ptype2->func = func3;
ptype2->next = &ptype3;

ptype3->func = func4;
ptype3->next = NULL;

If func2() freed skb, and return NET_RX_DROP, what will happen?

PS: I still don't understand why we should skip the first step, and call first function on second cycle?