[PATCH 19/72] gpu: ipu-v3: Protect more CM reg access with IPU lock

Steve Longerbeam slongerbeam at gmail.com
Tue Nov 4 17:54:59 PST 2014


On 11/04/2014 09:51 AM, Philipp Zabel wrote:
> Hi Steve,
>
> Am Freitag, den 31.10.2014, 15:54 -0700 schrieb Steve Longerbeam:
>> Some cm_reg accesses were not being protected by the IPU spin lock.
>>
>> Signed-off-by: Steve Longerbeam <steve_longerbeam at mentor.com>
>> ---
>>  drivers/gpu/ipu-v3/ipu-common.c |   22 ++++++++++++++++++++--
>>  1 file changed, 20 insertions(+), 2 deletions(-)
>>
>> diff --git a/drivers/gpu/ipu-v3/ipu-common.c b/drivers/gpu/ipu-v3/ipu-common.c
>> index f707d25..d3af206 100644
>> --- a/drivers/gpu/ipu-v3/ipu-common.c
>> +++ b/drivers/gpu/ipu-v3/ipu-common.c
>> @@ -46,11 +46,16 @@ static inline void ipu_cm_write(struct ipu_soc *ipu, u32 value, unsigned offset)
>>  
>>  void ipu_srm_dp_sync_update(struct ipu_soc *ipu)
>>  {
>> +	unsigned long flags;
>>  	u32 val;
>>  
>> +	spin_lock_irqsave(&ipu->lock, flags);
>> +
>>  	val = ipu_cm_read(ipu, IPU_SRM_PRI2);
>>  	val |= 0x8;
>>  	ipu_cm_write(ipu, val, IPU_SRM_PRI2);
>> +
>> +	spin_unlock_irqrestore(&ipu->lock, flags);
>>  }
>>  EXPORT_SYMBOL_GPL(ipu_srm_dp_sync_update);
> This is the only place this register touched, only bit 3 is ever enabled
> by software. This lock is not needed.

yes, I verified that too. Never mind then.

>> @@ -451,8 +456,17 @@ int ipu_idmac_get_current_buffer(struct ipuv3_channel *channel)
>>  {
>>  	struct ipu_soc *ipu = channel->ipu;
>>  	unsigned int chno = channel->num;
>> +	unsigned long flags;
>> +	int ret;
>> +
>> +	spin_lock_irqsave(&ipu->lock, flags);
>>  
>> -	return (ipu_cm_read(ipu, IPU_CHA_CUR_BUF(chno)) & idma_mask(chno)) ? 1 : 0;
>> +	ret = (ipu_cm_read(ipu, IPU_CHA_CUR_BUF(chno)) & idma_mask(chno)) ?
>> +		1 : 0;
>> +
>> +	spin_unlock_irqrestore(&ipu->lock, flags);
>> +
>> +	return ret;
> Dito. This register isn't written partially multiple times under the
> spinlock anywhere, so there is no gain from this lock around a register
> read.

Well, I was thinking get_current_buffer could race with
reset_current_buffer, but these functions are doing plain
readl() and writel() (i.e. not read-modify-write), so they
are atomic operations. So never mind this either.

>>  }
>>  EXPORT_SYMBOL_GPL(ipu_idmac_get_current_buffer);
>>  
>> @@ -569,10 +583,14 @@ EXPORT_SYMBOL_GPL(ipu_idmac_wait_busy);
>>  
>>  int ipu_wait_interrupt(struct ipu_soc *ipu, int irq, int ms)
>>  {
>> -	unsigned long timeout;
>> +	unsigned long flags, timeout;
>>  
>>  	timeout = jiffies + msecs_to_jiffies(ms);
>> +
>> +	spin_lock_irqsave(&ipu->lock, flags);
>>  	ipu_cm_write(ipu, BIT(irq % 32), IPU_INT_STAT(irq / 32));
>> +	spin_unlock_irqrestore(&ipu->lock, flags);
>> +
>>  	while (!(ipu_cm_read(ipu, IPU_INT_STAT(irq / 32) & BIT(irq % 32)))) {
>>  		if (time_after(jiffies, timeout))
>>  			return -ETIMEDOUT;
> This is a write to clear register, the only other place it is accessed
> is by the regmap irq handler. Can you think of a scenario where this
> lock would protect anything?

In this case I do see potential problems. What if thread A calls
ipu_wait_interrupt() while thread B is currently looping through
irq status bits in ipu_irq_handle()? A has now cleared an irq status
before B has had a chance to get to that status bit. Thus the irq
status is missed and the irq not handled. That being said this patch
will not fix the problem because ipu_wait_interrupt() and ipu_irq_handle()
are under different locks.

Maybe the solution is to verify ipu_wait_interrupt() is never called
on enabled irqs so that it doesn't interfere with irq handling. But I
don't know that that is true, is it?

Steve




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