Subject: [PATCH v3 06/13] irqchip: GICv3: ITS: LPI allocator

From: [email-masked] (Marc Zyngier)
Date: Mon, 24 Nov 2014 15:32:41 +0000

On 24/11/14 14:57, Jiang Liu wrote:
> 
> 
> On 2014/11/24 22:35, Marc Zyngier wrote:
>> LPIs are the type of interrupts that are used by the ITS. Given
>> the size of the namespace (anywhere between 16 and 32bit), interrupt
>> IDs are allocated in chunks of 32.
>>
>> Signed-off-by: Marc Zyngier <[email-masked]>
>> ---
>>  drivers/irqchip/irq-gic-v3-its.c | 103 +++++++++++++++++++++++++++++++++++++++
>>  1 file changed, 103 insertions(+)
>>
>> diff --git a/drivers/irqchip/irq-gic-v3-its.c b/drivers/irqchip/irq-gic-v3-its.c
>> index d24bebd..4154a16 100644
>> --- a/drivers/irqchip/irq-gic-v3-its.c
>> +++ b/drivers/irqchip/irq-gic-v3-its.c
>> @@ -586,3 +586,106 @@ static struct irq_chip its_irq_chip = {
>>  	.irq_eoi		= its_eoi_irq,
>>  	.irq_set_affinity	= its_set_affinity,
>>  };
>> +
>> +/*
>> + * How we allocate LPIs:
>> + *
>> + * The GIC has id_bits bits for interrupt identifiers. From there, we
>> + * must subtract 8192 which are reserved for SGIs/PPIs/SPIs. Then, as
>> + * we allocate LPIs by chunks of 32, we can shift the whole thing by 5
>> + * bits to the right.
> Just curious, why 32? sizeof(long) is 4 on ARM64?

No, sizeof(long) == 8, as on any sane 64bit architecture.

There are two reasons for this:
- the ID space is rather large (at least 16 bits, possibly 32 bits), so
we're trying not to allocate the whole bitmap in one go.
- 32 is the maximum a MSI-capable device can request. Allocating 32
interrupts in one go makes sure that these interrupts are contiguous and
satisfy the MSI requirements.

Hope this helps,

	M.
-- 
Jazz is not dead. It just smells funny...