[RFC PATCH v2] dmabuf-sync: Introduce buffer synchronization framework
Inki Dae
inki.dae at samsung.com
Thu Jun 20 04:15:57 PDT 2013
> -----Original Message-----
> From: Lucas Stach [mailto:l.stach at pengutronix.de]
> Sent: Thursday, June 20, 2013 7:11 PM
> To: Inki Dae
> Cc: 'Russell King - ARM Linux'; 'Inki Dae'; 'linux-fbdev'; 'YoungJun Cho';
> 'Kyungmin Park'; 'myungjoo.ham'; 'DRI mailing list'; linux-arm-
> kernel at lists.infradead.org; linux-media at vger.kernel.org
> Subject: Re: [RFC PATCH v2] dmabuf-sync: Introduce buffer synchronization
> framework
>
> Am Donnerstag, den 20.06.2013, 17:24 +0900 schrieb Inki Dae:
> [...]
> > > > In addition, please see the below more detail examples.
> > > >
> > > > The conventional way (without dmabuf-sync) is:
> > > > Task A
> > > > ----------------------------
> > > > 1. CPU accesses buf
> > > > 2. Send the buf to Task B
> > > > 3. Wait for the buf from Task B
> > > > 4. go to 1
> > > >
> > > > Task B
> > > > ---------------------------
> > > > 1. Wait for the buf from Task A
> > > > 2. qbuf the buf
> > > > 2.1 insert the buf to incoming queue
> > > > 3. stream on
> > > > 3.1 dma_map_sg if ready, and move the buf to ready queue
> > > > 3.2 get the buf from ready queue, and dma start.
> > > > 4. dqbuf
> > > > 4.1 dma_unmap_sg after dma operation completion
> > > > 4.2 move the buf to outgoing queue
> > > > 5. back the buf to Task A
> > > > 6. go to 1
> > > >
> > > > In case that two tasks share buffers, and data flow goes from Task A
> to
> > > Task
> > > > B, we would need IPC operation to send and receive buffers properly
> > > between
> > > > those two tasks every time CPU or DMA access to buffers is started
> or
> > > > completed.
> > > >
> > > >
> > > > With dmabuf-sync:
> > > >
> > > > Task A
> > > > ----------------------------
> > > > 1. dma_buf_sync_lock <- synpoint (call by user side)
> > > > 2. CPU accesses buf
> > > > 3. dma_buf_sync_unlock <- syncpoint (call by user side)
> > > > 4. Send the buf to Task B (just one time)
> > > > 5. go to 1
> > > >
> > > >
> > > > Task B
> > > > ---------------------------
> > > > 1. Wait for the buf from Task A (just one time)
> > > > 2. qbuf the buf
> > > > 1.1 insert the buf to incoming queue
> > > > 3. stream on
> > > > 3.1 dma_buf_sync_lock <- syncpoint (call by kernel side)
> > > > 3.2 dma_map_sg if ready, and move the buf to ready queue
> > > > 3.3 get the buf from ready queue, and dma start.
> > > > 4. dqbuf
> > > > 4.1 dma_buf_sync_unlock <- syncpoint (call by kernel side)
> > > > 4.2 dma_unmap_sg after dma operation completion
> > > > 4.3 move the buf to outgoing queue
> > > > 5. go to 1
> > > >
> > > > On the other hand, in case of using dmabuf-sync, as you can see the
> > > above
> > > > example, we would need IPC operation just one time. That way, I
> think we
> > > > could not only reduce performance overhead but also make user
> > > application
> > > > simplified. Of course, this approach can be used for all DMA device
> > > drivers
> > > > such as DRM. I'm not a specialist in v4l2 world so there may be
> missing
> > > > point.
> > > >
> > >
> > > You already need some kind of IPC between the two tasks, as I suspect
> > > even in your example it wouldn't make much sense to queue the buffer
> > > over and over again in task B without task A writing anything to it.
> So
> > > task A has to signal task B there is new data in the buffer to be
> > > processed.
> > >
> > > There is no need to share the buffer over and over again just to get
> the
> > > two processes to work together on the same thing. Just share the fd
> > > between both and then do out-of-band completion signaling, as you need
> > > this anyway. Without this you'll end up with unpredictable behavior.
> > > Just because sync allows you to access the buffer doesn't mean it's
> > > valid for your use-case. Without completion signaling you could easily
> > > end up overwriting your data from task A multiple times before task B
> > > even tries to lock the buffer for processing.
> > >
> > > So the valid flow is (and this already works with the current APIs):
> > > Task A Task B
> > > ------ ------
> > > CPU access buffer
> > > ----------completion signal--------->
> > > qbuf (dragging buffer into
> > > device domain, flush caches,
> > > reserve buffer etc.)
> > > |
> > > wait for device operation to
> > > complete
> > > |
> > > dqbuf (dragging buffer back
> > > into CPU domain, invalidate
> > > caches, unreserve)
> > > <---------completion signal------------
> > > CPU access buffer
> > >
> >
> > Correct. In case that data flow goes from A to B, it needs some kind
> > of IPC between the two tasks every time as you said. Then, without
> > dmabuf-sync, how do think about the case that two tasks share the same
> > buffer but these tasks access the buffer(buf1) as write, and data of
> > the buffer(buf1) isn't needed to be shared?
> >
> Sorry, I don't see the point you are trying to solve here. If you share
> a buffer and want its content to be clearly defined at every point in
> time you have to synchronize the tasks working with the buffer, not just
> the buffer accesses itself.
>
> Easiest way to do so is doing sync through userspace with out-of-band
> IPC, like in the example above.
In my opinion, that's not definitely easiest way. What I try to do is to avoid using *the out-of-band IPC*. As I mentioned in document file, the conventional mechanism not only makes user application complicated-user process needs to understand how the device driver is worked-but also may incur performance overhead by using the out-of-band IPC. The above my example may not be enough to you but there would be other cases able to use my approach efficiently.
> A more advanced way to achieve this
> would be using cross-device fences to avoid going through userspace for
> every syncpoint.
>
Ok, maybe there is something I missed. So question. What is the cross-device fences? dma fence?. And how we can achieve the synchronization mechanism without going through user space for every syncpoint; CPU and DMA share a same buffer?. And could you explain it in detail as long as possible like I did?
> >
> > With dmabuf-sync is:
> >
> > Task A
> > ----------------------------
> > 1. dma_buf_sync_lock <- synpoint (call by user side)
> > 2. CPU writes something to buf1
> > 3. dma_buf_sync_unlock <- syncpoint (call by user side)
> > 4. copy buf1 to buf2
> Random contents here? What's in the buffer, content from the CPU write,
> or from V4L2 device write?
>
Please presume that buf1 is physically non contiguous memory, and buf2 is physically contiguous memory; device A without IOMMU is seeing buf2. We would need to copy buf1 to buf2 to send the contents of the buf1 to device A because DMA of the device A cannot access the buf1 directly. And CPU and V4L2 device don't share the contents of the buf1 but share the buf1 as storage.
Thanks,
Inki Dae
> > 5. go to 1
> >
> >
> > Task B
> > ---------------------------
> > 1. dma_buf_sync_lock
> > 2. CPU writes something to buf3
> > 3. dma_buf_sync_unlock
> > 4. qbuf the buf3(src) and buf1(dst)
> > 4.1 insert buf3,1 to incoming queue
> > 4.2 dma_buf_sync_lock <- syncpoint (call by kernel side)
> > 5. stream on
> > 5.1 dma_map_sg if ready, and move the buf to ready queue
> > 5.2 get the buf from ready queue, and dma start.
> > 6. dqbuf
> > 6.1 dma_buf_sync_unlock <- syncpoint (call by kernel side)
> > 6.2 dma_unmap_sg after dma operation completion
> > 6.3 move the buf3,1 to outgoing queue
> > 7. go to 1
> >
>
> Regards,
> Lucas
> --
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