[PATCH][RFC] fsldma: fix performance degradation by optimizing spinlock use.
Ira W. Snyder
iws at ovro.caltech.edu
Wed Nov 23 05:59:25 EST 2011
On Tue, Nov 22, 2011 at 12:55:05PM +0800, b29237 at freescale.com wrote:
> From: Forrest Shi <b29237 at freescale.com>
>
> dma status check function fsl_tx_status is heavily called in
> a tight loop and the desc lock in fsl_tx_status contended by
> the dma status update function. this caused the dma performance
> degrades much.
>
> this patch releases the lock in the fsl_tx_status function.
> I believe it has no neglect impact on the following call of
> dma_async_is_complete(...).
>
> we can see below three conditions will be identified as success
> a) x < complete < use
> b) x < complete+N < use+N
> c) x < complete < use+N
> here complete is the completed_cookie, use is the last_used
> cookie, x is the querying cookie, N is MAX cookie
>
> when chan->completed_cookie is being read, the last_used may
> be incresed. Anyway it has no neglect impact on the dma status
> decision.
>
> Signed-off-by: Forrest Shi <xuelin.shi at freescale.com>
> ---
> drivers/dma/fsldma.c | 5 -----
> 1 files changed, 0 insertions(+), 5 deletions(-)
>
> diff --git a/drivers/dma/fsldma.c b/drivers/dma/fsldma.c
> index 8a78154..1dca56f 100644
> --- a/drivers/dma/fsldma.c
> +++ b/drivers/dma/fsldma.c
> @@ -986,15 +986,10 @@ static enum dma_status fsl_tx_status(struct dma_chan *dchan,
> struct fsldma_chan *chan = to_fsl_chan(dchan);
> dma_cookie_t last_complete;
> dma_cookie_t last_used;
> - unsigned long flags;
> -
> - spin_lock_irqsave(&chan->desc_lock, flags);
>
This will cause a bug. See below for a detailed explanation. You need
this instead:
/*
* On an SMP system, we must ensure that this CPU has seen the
* memory accesses performed by another CPU under the
* chan->desc_lock spinlock.
*/
smp_mb();
> last_complete = chan->completed_cookie;
> last_used = dchan->cookie;
>
> - spin_unlock_irqrestore(&chan->desc_lock, flags);
> -
> dma_set_tx_state(txstate, last_complete, last_used, 0);
> return dma_async_is_complete(cookie, last_complete, last_used);
> }
Facts:
- dchan->cookie is the same member as chan->common.cookie (same memory location)
- chan->common.cookie is the "last allocated cookie for a pending transaction"
- chan->completed_cookie is the "last completed transaction"
I have replaced "dchan->cookie" with "chan->common.cookie" in the below
explanation, to keep everything referenced from the same structure.
Variable usage before your change. Everything is used locked.
- RW chan->common.cookie (fsl_dma_tx_submit)
- R chan->common.cookie (fsl_tx_status)
- R chan->completed_cookie (fsl_tx_status)
- W chan->completed_cookie (dma_do_tasklet)
Variable usage after your change:
- RW chan->common.cookie LOCKED
- R chan->common.cookie NO LOCK
- R chan->completed_cookie NO LOCK
- W chan->completed_cookie LOCKED
What if we assume that you have a 2 CPU system (such as a P2020). After
your changes, one possible sequence is:
=== CPU1 - allocate + submit descriptor: fsl_dma_tx_submit() ===
spin_lock_irqsave
descriptor->cookie = 20 (x in your example)
chan->common.cookie = 20 (used in your example)
spin_unlock_irqrestore
=== CPU2 - immediately calls fsl_tx_status() ===
chan->common.cookie == 19
chan->completed_cookie == 19
descriptor->cookie == 20
Since we don't have locks anymore, CPU2 may not have seen the write to
chan->common.cookie yet.
Also assume that the DMA hardware has not started processing the
transaction yet. Therefore dma_do_tasklet() has not been called, and
chan->completed_cookie has not been updated.
In this case, dma_async_is_complete() (on CPU2) returns DMA_SUCCESS,
even though the DMA operation has not succeeded. The DMA operation has
not even started yet!
The smp_mb() fixes this, since it forces CPU2 to have seen all memory
operations that happened before CPU1 released the spinlock. Spinlocks
are implicit SMP memory barriers.
Therefore, the above example becomes:
smp_mb();
chan->common.cookie == 20
chan->completed_cookie == 19
descriptor->cookie == 20
Then dma_async_is_complete() returns DMA_IN_PROGRESS, which is correct.
Thanks,
Ira
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