Hi,
Le Wed, 29 Aug 2007 11:10:37 +0200,
Johannes Schmid <johnny@xxxxxxxx> a écrit :
> spin_lock(&task->mm->page_table_lock);
>
> pte = pte_offset_map_lock(task->mm, pmd, addr, &ptl);
>
> spin_unlock(&task->mm->page_table_lock);
pte_offset_map_lock() does:
932 #define pte_offset_map_lock(mm, pmd, address, ptlp) \
933 ({ \
934 spinlock_t *__ptl = pte_lockptr(mm, pmd); \
935 pte_t *__pte = pte_offset_map(pmd, address); \
936 *(ptlp) = __ptl; \
937 spin_lock(__ptl); \
938 __pte; \
939 })
So it takes the lock returned by pte_lockptr, which is defined as
follows:
929 #define pte_lockptr(mm, pmd) ({(void)(pmd);
&(mm)->page_table_lock;})
Which takes the page_table_lock you've already taken previously.
So I'd say that whenever you reach your call to pte_offset_map_lock(),
your kernel will deadlock.
Looking more closely at the code, you can see that there are cases
where pte_offset_map_lock() doesn't take the mm->page_table_lock, but
page->ptl. It's when you have NR_CPUS >= CONFIG_SPLIT_PTLOCK_CPUS. See
http://users.sosdg.org/~qiyong/lxr/source/include/linux/mm.h#L910.
However, CONFIG_SPLIT_PTLOCK_CPUS is defined as follows:
config SPLIT_PTLOCK_CPUS
int
default "4096" if ARM && !CPU_CACHE_VIPT
default "4096" if PARISC && !PA20
default "4"
So on i386, it defaults to 4. So unless you have more than 4
processors, pte_offset_map_lock() takes mm->page_table_lock.
Sincerly,
Thomas
--
Thomas Petazzoni - thomas.petazzoni@xxxxxxxx
http://{thomas,sos,kos}.enix.org - http://www.toulibre.org
http://www.{livret,agenda}dulibre.org
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