Messages in this thread | | | From | "Serge Kuznetsov" <> | Subject | Re: [NET] Possible bug in netif_receive_skb | Date | Thu, 14 Nov 2002 15:15:26 -0500 |
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> > ->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? - To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/
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