Re: [PATCH v7 00/11] kallsyms: Optimizes the performance of lookup symbols

["Leizhen (ThunderTown)" <thunder.leizhen@huawei.com>]

On 2022/10/29 20:49, David Laight wrote:
>>>> On 2022/10/27 3:03, Luis Chamberlain wrote:
>>>>> On Wed, Oct 26, 2022 at 02:44:36PM +0800, Leizhen (ThunderTown) wrote:
>>>>>> On 2022/10/26 1:53, Luis Chamberlain wrote:
>>>>>>> This answers how we don't use a hash table, the question was *should* we
>>>>>>> use one?
> 
> (Probably brainfart) thought...
> 
> Is the current table (effectively) a sorted list of strings?
> So the lookup is a binary chop - so O(log(n)).

Currently not sorted.

> 
> But your hashes are having 'trouble' stopping one chain
> being very long?
> So a linear search of that hash chain is slow.
> In fact that sort of hashed lookup in O(n).

You've analyzed it very well. The hash method is not good for sorting names
and then searching in binary mode. I figured it out when I was working on
the design these days.

Current Implementation:
---------------------------------------
| idx | addresses | markers |  names  |
---------------------------------------
|  0  |    addr0  |         |  name0  |
|  1  |    addr1  |         |  name1  |
| ... |    addrx  |   [0]   |  namex  |
| 255 |    addrx  |         |  name255|
---------------------------------------
| 256 |  addr256  |         |  name256|
| ... |    addrx  |   [1]   |  namex  |
| 511 |  addr511  |         |  name511|
---------------------------------------

markers[0] = offset_of(name0)
markers[1] = offset_of(name256)

1. Find name by address
   binary search addresses[], get idx, traverse names[] from  markers[idx>>8] to markers[(idx>>8) + 1], return name

2. Find address by name
   traverse names[], get idx, return addresses[idx]

Hash Implementation:
Add two new arrays: hash_table[] and names_offsets[]

-----------------------------------------------------------------
| key |      hash_table       |         names_offsets           |
|---------------------------------------------------------------|
|  0  |  names_offsets[key=0] | offsets of all names with key=0 |
|  1  |  names_offsets[key=1] | offsets of all names with key=1 |
| ... |          ...          | offsets of all names with key=k |
|---------------------------------------------------------------|

hash_table[0] = 0
hash_table[1] = hash_table[0] + sizeof(names_offsets[0]) * number_of_names(key=0)
hash_table[2] = hash_table[1] + sizeof(names_offsets[0]) * number_of_names(key=1)

1. Find address by name
   hash name, get key, traverse names_offsets[] from index=hash_table[key] to
   index=hash_table[key+1], get the offset of name in names[]. binary search markers[],
   get index, then traverse names[] from  markers[index] to markers[index + 1], until
   match the offset of name, return addresses[idx].
2. Find address by name
   No change.

Sorted names Implementation:
Add two new arrays: offsets_of_addr_to_name[] and offsets_of_name[]

offsets_of_addr_to_name[i] = offset of name i in names[]
offsets_of_name[i]         = offset of sorted name i in names[]

1. Find name by address
   binary search addresses[], get idx, return names[offsets_of_addr_to_name[idx]]

2. Find address by name
   binary search offsets_of_name[], get idx, return addresses[idx]

> 
> What if the symbols were sorted by hash then name?
> (Without putting the hash into each entry.)
> Then the code could do a binary chop search over
> the symbols with the same hash value.
> The additional data is then an array of symbol numbers
> indexed by the hash - 32 bits for each bucket.
> 
> If the hash table has 0x1000 entries it saves 12 compares.
> (All of which are likely to be data cache misses.)
> 
> If you add the hash to each table entry then you can do
> a binary chop search for the hash itself.
> While this is the same search as is done for the strings
> the comparison (just a number) will be faster than a
> string compare.
> 
> 	David
> 
> -
> Registered Address Lakeside, Bramley Road, Mount Farm, Milton Keynes, MK1 1PT, UK
> Registration No: 1397386 (Wales)
> 

-- 
Regards,
  Zhen Lei