Re: [PATCH 2/2] dma: mmp_pdma: clear DRCMR when free a phy channel

[Andy Shevchenko]

On Mon, Jun 3, 2013 at 11:02 AM, Xiang Wang <wangxfdu@xxxxxxxxx> wrote:
> From: Xiang Wang <wangx@xxxxxxxxxxx>
>
> In mmp pdma, phy channels are allocated/freed dynamically.
> The mapping from DMA request to DMA channel number in DRCMR
> should be cleared when a phy channel is freed. Otherwise
> conflicts will happen when:
> 1. A is using channel 2 and free it after finished, but A
> still maps to channel 2 in DRCMR of A.
> 2. Now another one B gets channel 2. So B maps to channel 2
> too in DRCMR of B.
> In the datasheet, it is described that "Do not map two active
> requests to the same channel since it produces unpredictable
> results" and we can observe that during test.


> --- a/drivers/dma/mmp_pdma.c
> +++ b/drivers/dma/mmp_pdma.c
> @@ -252,9 +252,15 @@ static void free_phy(struct mmp_pdma_chan *pchan)
>  {
>         struct mmp_pdma_device *pdev = to_mmp_pdma_dev(pchan->chan.device);
>         unsigned long flags;
> +       u32 reg;

May be empty line here (and maybe even in previous patch)?

>         if (!pchan->phy)
>                 return;
>
> +       /* clear the channel mapping in DRCMR */
> +       reg = pchan->phy->vchan->drcmr;
> +       reg = (((reg) < 64) ? 0x0100 : 0x1100) + (((reg) & 0x3f) << 2);

Too many braces.
It is not a macro, you don't need to embrace reg.

> +       writel(0, pchan->phy->base + reg);
> +
>         spin_lock_irqsave(&pdev->phy_lock, flags);
>         pchan->phy->vchan = NULL;
>         pchan->phy = NULL;



--
With Best Regards,
Andy Shevchenko
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