Re: [PATCH v2] i2c: designware: Fix corrupted memory seen in the ISR

From: Jan Bottorff
Date: Wed Sep 20 2023 - 15:14:59 EST


On 9/20/2023 6:27 AM, Yann Sionneau wrote:
Hi,

On 20/09/2023 11:08, Wolfram Sang wrote:
same thread." [1] Thus I'd suggest the next fix for the problem:

--- a/drivers/i2c/busses/i2c-designware-common.c
+++ b/drivers/i2c/busses/i2c-designware-common.c
@@ -72,7 +72,10 @@ static int dw_reg_write(void *context, unsigned int reg, unsigned int val)
  {
      struct dw_i2c_dev *dev = context;
-    writel_relaxed(val, dev->base + reg);
+    if (reg == DW_IC_INTR_MASK)
+        writel(val, dev->base + reg);
+    else
+        writel_relaxed(val, dev->base + reg);
      return 0;
  }

(and similar changes for dw_reg_write_swab() and dw_reg_write_word().)

What do you think?
To me, this looks reasonable and much more what I would have expected as
a result (from a high level point of view). Let's hope it works. I am
optimistic, though...

It works if we make sure all the other register accesses to the designware i2c IP can't generate IRQ.

Meaning that all register accesses that can trigger an IRQ are enclosed in between a call to i2c_dw_disable_int() and a call to regmap_write(dev->map, DW_IC_INTR_MASK, DW_IC_INTR_MASTER_MASK); or equivalent.

It seems to be the case, I'm not sure what's the best way to make sure it will stay that way.

Moreover, maybe writes to IC_ENABLE register should also use the non-relaxed writel() version?

Since one could do something like:

[ IP is currently disabled ]

1/ enable interrupts in DW_IC_INTR_MASK

2/ update some variable in dev-> structure in DDR

3/ enable the device by writing to IC_ENABLE, thus triggering for instance the TX_FIFO_EMPTY irq.


It does seem like there are a variety of register write combinations that could immediately cause an interrupt, so would need a barrier.

I understand barriers a bit better now, although still wonder about some cases. Like say you write to some driver memory and then write the DW command fifo register, and it does not immediately cause an interrupt, but will in the future. Even without the barrier the memory write would typically become visible to other cores after some small amount of time, but don't see that's it's architecturally guaranteed to be visible. This implies the barrier is perhaps needed on many/all of the registers.

Jan