Re: [PATCH 4/5] locking/lockdep: Reuse free chain_hlocks entries

[Waiman Long <longman@redhat.com>]

On 12/13/19 5:50 AM, Peter Zijlstra wrote:
> On Fri, Dec 13, 2019 at 11:25:25AM +0100, Peter Zijlstra wrote:
>
>> void push_block(struct chain_block **bp, struct chain_block *b)
>> {
>> 	b->next = *bp;
>> 	*bp = b;
>> }
>>
>> /* could contemplate ilog2() buckets */
>> int size2bucket(int size)
>> {
>> 	return size >= MAX_BUCKET ? 0 : size;
>> }
> If you make the allocation granularity 4 entries, then you can have
> push_block() store the chain_block * at the start of every free entry,
> which would enable merging adjecent blocks.
>
> That is, if I'm not mistaken these u16 chain entries encode the class
> index (per lock_chain_get_class()). And since the lock_classes array is
> MAX_LOCKDEP_KEYS, or 13 bits, big, we have the MSB of each entry spare.
>
> union {
> 	struct {
> 		u16 hlock[4];
> 	}
> 	u64 addr;
> } ponies;
>
> So if we make the rule that:
>
> 	!(idx % 4) && (((union ponies *)chain_hlock[idx])->addr & BIT_ULL(63))
>
> encodes a free block at said addr, then we should be able to detect if:
>
> 	chain_hlock[b->base + b->size]
>
> is also free and merge the blocks.
>
> (we just have to fix up the MSB of the address, not all arches have
> negative addresses for kernel space)
>
That is an interesting idea. It will eliminate the need of a separate
array to track the free chain_hlocks. However, if there are n chains
available, it will waste about 3n bytes of storage, on average.

I have a slightly different idea. I will enforce a minimum allocation
size of 2. For a free block, the first 2 hlocks for each allocation
block will store a 32-bit integer (hlock[0] << 16)|hlock[1]:

Bit 31: always 1
Bits 24-30: block size
Bits 0-23: index to the next free block.

In this way, the wasted space will be k bytes where k is the number of
1-entry chains. I don't think merging adjacent blocks will be that
useful at this point. We can always add this capability later on if it
is found to be useful.

Cheers,
Longman