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Re: [dm-devel] possible dm-crypt leak over loop

> > > Can anybody else confirm this?
> > 
> > The original report came from Steffen Michalke (CC'd), but I find it
> > very weird that you can't trigger the lockup. It takes about a second
> > here on a UP non-preempt box to lock up.
> Perhaps my machine has too much memory. ;)
> > Why I think it's a memory leak is due to the sysrq-Meminfo trace
> > and the call-trace showing 
> > crypt_alloc_buffer => mempool_alloc => io_schedule
> That's normal. For write-outs dm-crypt has to allocate some buffers
> where it can put the encrypted data. mempool_alloc means that it ran out
> of preallocated buffers and waits for some in-flight I/O to finish which
> should free some buffers.
> It looks like this is a dm-crypt <-> loop deadlock. The problem with
> loop is that it goes back through the page cache. The loop thread is
> also waiting for memory to actually be able to write out some more
> pages.
> So this seems to be a loop bug. Loop is waiting for pages to become free
> to actually be able to write some pages. dm-crypt uses a mempool to
> avoid this deadlock but if the underlying device deadlocks it can't do
> anything about that.
> loop does a: alloc_page(GFP_NOIO | __GFP_HIGHMEM);
> There is no guarantee that this will succeed. It probably should also
> have some sort of emergency pool.

Ok, but - this is a late revision, the allocation you pointed out did
not exist in for example 2.6.10. I tried this my test script on a 128MB
machine a few kernels back and it did happen in 2.6.10, maybe not as
fast however.

The place where loop does page allocations in writeout in 2.6.10 is 

And the grab_cache_page looks like a common filesystem called function,
but I guess the difference is that this time the data to writeout is in
kernel memory and not user-space owned. I figure loop shouldn't call
this function that does bare alloc_page, but needs something that uses a
mempool to prevent it from deadlocks when stacking with other drivers.

I don't think I've got the whole picture here, but I'm grateful for your
previous explanation, it helps alot!


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