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Date:	Tue, 18 Mar 2008 00:30:35 +0100 (CET)
From:	Jan Engelhardt <jengelh@...putergmbh.de>
To:	Linux Kernel Mailing List <linux-kernel@...r.kernel.org>
Subject: vfree with spin_lock_bh

Hi,


while transforming some code with big allocations (like 120 KB) from 
kmalloc to vmalloc — virtual contiguity is sufficient — I hit a
BUG_ON in mm/vmalloc.c a number of times:

	void vfree(const void *addr)
	{
	        BUG_ON(in_interrupt());
	        __vunmap(addr, 1);
	}

First I was thinking “how could iptables -F run in interrupt context?”,
but apparently, it does seem to make a difference:

	...
	spin_lock_bh(&a_local_spinlock);
	list_del_rcu(&node->list);
	printk(KERN_INFO "Interrupt? %lu\n", in_interrupt());
	/* vfree not worky here */
	spin_unlock_bh(&a_local_spinlock);
	printk(KERN_INFO "Interrupt? %lu\n", in_interrupt());
	/* now possible */
	vfree(node);
	...

and this gives (x86_32)

	Interrupt? 256
	Interrupt? 0

So this may be a "property" of spinlocks, but it is a bit strange to me.
Why should not I be able to call vfree() when I am, in fact, in
user context (but with a bh spinlock held...).

Do I perhaps need a non-bh spinlock? There's RCU going on on that 
linked list so I am not sure whether I could just call the normal 
spin_lock() function.

Looking at the code of _spin_lock_bh in kernel/spinlock.c reveals that 
it is actually disabling preempt instead of being in an interrupt. 
Making an uneducated guess, would

	BUG_ON(in_interrupt() != 0 && in_interrupt() != 256)

in vfree() be safe?


thanks,
Jan
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