Messages in this thread | | | Date | Tue, 1 Mar 2011 10:51:43 +0100 (CET) | From | Thomas Gleixner <> | Subject | Re: [PATCH v6 1/1] PRUSS UIO driver support |
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On Tue, 1 Mar 2011, Pratheesh Gangadhar wrote: > + > +static spinlock_t lock;
static DEFINE_SPINLOCK(lock);
> +static struct clk *pruss_clk; > +static struct uio_info *info; > +static dma_addr_t sram_paddr, ddr_paddr; > +static void *prussio_vaddr, *sram_vaddr, *ddr_vaddr; > + > +static irqreturn_t pruss_handler(int irq, struct uio_info *dev_info) > +{ > + unsigned long flags; > + int val, intr_mask = (1 << (irq - 1)); > + void __iomem *base = dev_info->mem[0].internal_addr; > + void __iomem *intren_reg = base + PINTC_HIER; > + void __iomem *intrstat_reg = base + PINTC_HIPIR + ((irq - 1) << 2); > + > + spin_lock_irqsave(&lock, flags);
spin_lock() is enough as we run handlers with interrupts disabled.
> + val = ioread32(intren_reg); > + /* Is interrupt enabled and active ? */ > + if (!(val & intr_mask) && (ioread32(intrstat_reg) & HIPIR_NOPEND)) { > + spin_unlock_irqrestore(&lock, flags); > + return IRQ_NONE; > + } > + > + /* Disable interrupt */ > + iowrite32((val & ~intr_mask), intren_reg); > + spin_unlock_irqrestore(&lock, flags); > + return IRQ_HANDLED; > +}
So now you still have not solved the problem of user space enabling an interrupt again. That's racy as well and you can solve it by providing an uio->irq_control function which handles the interrupt enable register under the lock as well.
Thanks,
tglx
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