On Mon, 2007-03-12 at 17:32 -0700, Christopher Li wrote: > That is because sparse can't distinguish which lock is > acquired. It is actually hard to get that information. > Even though the expression is the same, the actual lock > might be different. > > e.g. > > redlock(foo->redlock); > foo = bar; > redunlock (foo->redlock); Just an idea. The Linux "runtime locking correctness validator" (see Documentation/lockdep-design.txt in the Linux sources) distinguishes between lock classes. Sparse could use a similar approach. I think it would not catch your example, but it would catch a more realistic case when one lock is acquired and another is released even by the same function with the same attributes: lock(foo->redlock); lock(foo->bluelock); unlock(foo->redlock); unlock(foo->bluelock); Lock class could be just a unique reference to a place where the lock was declared. That would put some limitations on what a lock could be (an address of a variable or a field), but I think it's OK. -- Regards, Pavel Roskin - To unsubscribe from this list: send the line "unsubscribe linux-sparse" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
