On Fri, 2014-02-28 at 11:36 +0100, Maxime Ripard wrote:
> Hi Andy,
>
> On Tue, Feb 25, 2014 at 01:28:15PM +0200, Andy Shevchenko wrote:
> > > +static irqreturn_t sun6i_dma_interrupt(int irq, void *dev_id)
> > > +{
> > > + struct sun6i_dma_dev *sdev = (struct sun6i_dma_dev *)dev_id;
> > > + struct sun6i_vchan *vchan;
> > > + struct sun6i_pchan *pchan;
> > > + int i, j, ret = 0;
> > > + u32 status;
> > > +
> > > + for (i = 0; i < 2; i++) {
> > > + status = readl(sdev->base + DMA_IRQ_STAT(i));
> > > + if (!status) {
> > > + ret |= IRQ_NONE;
> >
> > Maybe move this to definition block.
> >
> > > + continue;
> > > + }
> > > +
> > > + dev_dbg(sdev->slave.dev, "DMA irq status %s: 0x%x\n",
> > > + i ? "high" : "low", status);
> > > +
> > > + writel(status, sdev->base + DMA_IRQ_STAT(i));
> > > +
> > > + for (j = 0; (j < 8) && status; j++) {
> > > + if (status & DMA_IRQ_QUEUE) {
> > > + pchan = sdev->pchans + j;
> > > + vchan = pchan->vchan;
> > > +
> > > + if (vchan) {
> > > + unsigned long flags;
> > > +
> > > + spin_lock_irqsave(&vchan->vc.lock,
> > > + flags);
> > > + vchan_cookie_complete(&pchan->desc->vd);
> > > + pchan->done = pchan->desc;
> > > + spin_unlock_irqrestore(&vchan->vc.lock,
> > > + flags);
> > > + }
> > > + }
> > > +
> > > + status = status >> 4;
> > > + }
> > > +
> > > + ret |= IRQ_HANDLED;
> >
> > In case one is handled, another is not, what you have to do?
>
> The interrupt status is split across two registers. In the case where
> one of the two register reports an interrupt, we still have to handle
> our interrupt, we actually did, so we have to return IRQ_HANDLED.
You removed the code below this assignment, but if I remember correctly
you check for exact value there.
In case of one is not handled and the other is handled you will have ret
= IRQ_HANDLED | IRQ_NONE. Thus, your following code will not be
executed. Is it by design?
--
Andy Shevchenko <andriy.shevchenko@xxxxxxxxxxxxxxx>
Intel Finland Oy
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