>
> ->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?
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This archive was generated by hypermail 2b29 : Fri Nov 15 2002 - 22:00:34 EST