dma: mmp_pdma: clear DRCMR when free a phy channel
authorXiang Wang <wangx@marvell.com>
Tue, 18 Jun 2013 06:55:59 +0000 (14:55 +0800)
committerVinod Koul <vinod.koul@intel.com>
Mon, 5 Aug 2013 04:02:27 +0000 (09:32 +0530)
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.

Signed-off-by: Xiang Wang <wangx@marvell.com>
Signed-off-by: Vinod Koul <vinod.koul@intel.com>
drivers/dma/mmp_pdma.c

index b1a260002cf88328cf9d9e023eb3fcc3d96b5a16..9a32f2d174b0bc549ce79a33d4ed962f19c71094 100644 (file)
@@ -252,10 +252,16 @@ static void mmp_pdma_free_phy(struct mmp_pdma_chan *pchan)
 {
        struct mmp_pdma_device *pdev = to_mmp_pdma_dev(pchan->chan.device);
        unsigned long flags;
+       u32 reg;
 
        if (!pchan->phy)
                return;
 
+       /* clear the channel mapping in DRCMR */
+       reg = pchan->phy->vchan->drcmr;
+       reg = ((reg < 64) ? 0x0100 : 0x1100) + ((reg & 0x3f) << 2);
+       writel(0, pchan->phy->base + reg);
+
        spin_lock_irqsave(&pdev->phy_lock, flags);
        pchan->phy->vchan = NULL;
        pchan->phy = NULL;