linux-kernel - Re: perf: use-after-free in perf

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Date:   Thu, 16 Mar 2017 14:57:35 +0100
From:   Peter Zijlstra <peterz@...radead.org>
To:     Oleg Nesterov <oleg@...hat.com>
Cc:     Dmitry Vyukov <dvyukov@...gle.com>, Ingo Molnar <mingo@...hat.com>,
        Arnaldo Carvalho de Melo <acme@...nel.org>,
        Alexander Shishkin <alexander.shishkin@...ux.intel.com>,
        LKML <linux-kernel@...r.kernel.org>,
        Mathieu Desnoyers <mathieu.desnoyers@...icios.com>,
        syzkaller <syzkaller@...glegroups.com>
Subject: Re: perf: use-after-free in perf_release

On Wed, Mar 15, 2017 at 05:43:02PM +0100, Oleg Nesterov wrote:
> 	static inline int list_is_first(const struct list_head *list,
> 					const struct list_head *head)
> 	{
> 		return head->next == list;
> 	}
> 
> won't be symmetrical with list_is_last() we already have.

This is the one that makes sense to me though; that is, the current
list_is_last() doesn't make sense to me.

I would expect:

static inline int list_is_last(const struct list_head *list,
				const struct list_head *head)
{
	return head->prev == list
}

because @head is the list argument (yes, I know, horrible naming!).