Message-ID: <ca8763c6-6bac-47fb-4b1e-fa4e88e2c422@loongson.cn>
Date:   Fri, 10 Feb 2023 15:12:58 +0800
From:   maobibo <maobibo@...ngson.cn>
To:     Huacai Chen <chenhuacai@...nel.org>,
        David Laight <David.Laight@...lab.com>
Cc:     WANG Xuerui <kernel@...0n.name>,
        Jiaxun Yang <jiaxun.yang@...goat.com>,
        "loongarch@...ts.linux.dev" <loongarch@...ts.linux.dev>,
        "linux-kernel@...r.kernel.org" <linux-kernel@...r.kernel.org>
Subject: Re: [PATCH v2] LoongArch: add checksum optimization for 64-bit system



在 2023/2/10 11:21, Huacai Chen 写道:
> This commit comes from the old internal kernel, I want to know which
> one has better performance.
> 
> https://github.com/loongson/linux/commit/92a6df48ccb73dd2c3dc1799add08adf0e0b0deb

There is no obvious performance difference between asm code csum_partial function
and uint128 c code. Tested with buffer size 1500/4096, uint128 c code method is
about 5% faster than asm code csum_partial.

regards
bibo, mao
> 
> On Thu, Feb 9, 2023 at 8:39 PM David Laight <David.Laight@...lab.com> wrote:
>>
>> From: maobibo
>>> Sent: 09 February 2023 11:55
>>>
>>>
>>> 在 2023/2/9 17:35, David Laight 写道:
>>>> From: Bibo Mao
>>>>> Sent: 09 February 2023 03:59
>>>>>
>>>>> loongArch platform is 64-bit system, which supports 8 bytes memory
>>>>> accessing, generic checksum function uses 4 byte memory access.
>>>>> This patch adds 8-bytes memory access optimization for checksum
>>>>> function on loongArch. And the code comes from arm64 system.
>>>>
>>>> How fast do these functions actually run (in bytes/clock)?
>>> With uint128 method, there will unrolled loop, instruction
>>> can execute in parallel. It gets the best result on loongarch
>>> system where there is no neither carry flag nor post-index
>>> addressing modes.
>>
>> We're probably almost agreeing...
>>
>>> Here is the piece of disassemble code with uint128 method:
>>
>> Load 8 values:
>>
>>>    120000a40:   28c0222f        ld.d    $r15,$r17,8(0x8)
>>>    120000a44:   28c0622a        ld.d    $r10,$r17,24(0x18)
>>>    120000a48:   28c0a230        ld.d    $r16,$r17,40(0x28)
>>>    120000a4c:   28c0e232        ld.d    $r18,$r17,56(0x38)
>>>    120000a50:   28c0022e        ld.d    $r14,$r17,0
>>>    120000a54:   28c0422d        ld.d    $r13,$r17,16(0x10)
>>>    120000a58:   28c0822b        ld.d    $r11,$r17,32(0x20)
>>>    120000a5c:   28c0c22c        ld.d    $r12,$r17,48(0x30)
>>
>> Pairwise add them
>>
>>>    120000a60:   0010b9f7        add.d   $r23,$r15,$r14
>>>    120000a64:   0010b54d        add.d   $r13,$r10,$r13
>>>    120000a68:   0010b24c        add.d   $r12,$r18,$r12
>>>    120000a6c:   0010ae0b        add.d   $r11,$r16,$r11
>>
>> Generate 4 'carry' bits
>>
>>>    120000a70:   0012c992        sltu    $r18,$r12,$r18
>>>    120000a74:   0012beee        sltu    $r14,$r23,$r15
>>>    120000a78:   0012c170        sltu    $r16,$r11,$r16
>>>    120000a7c:   0012a9aa        sltu    $r10,$r13,$r10
>>
>> Add the carry bits onto the sums.
>> I've not quite worked out which add is which!
>> But I think you've missed a few adds here.
>>
>>>    120000a80:   0010ae0f        add.d   $r15,$r16,$r11
>>>    120000a84:   0010ddce        add.d   $r14,$r14,$r23
>>>    120000a88:   0010b250        add.d   $r16,$r18,$r12
>>>    120000a8c:   0010b54d        add.d   $r13,$r10,$r13
>>>    120000a90:   0010b5d2        add.d   $r18,$r14,$r13
>>>    120000a94:   0010c1f0        add.d   $r16,$r15,$r16
>>
>> Somewhere each value needs an add, an sltu to generate the 'carry',
>> and an add for the carry itself.
>> If you sum the carry bits into a separate register it is
>> possible to get a both adds and the sltu (for different values)
>> to run in the same clock (on a suitable cpu).
>> If there are 4 integer units you can also get the loop instructions
>> 'for free' and unrolling 8 times may not be needed at all.
>>
>> ...
>>> There is no post-index addressing modes on loongarch,
>>>       val = *mem;  // 64bit read
>>>         mem++;
>>>       sum += val;
>>>       carry = sum < val;
>>>       carry_sum += carry;
>>> it takes 5 instruction and these 5 instructions depends on previous instr.
>>
>> I'd assume the loop was unrolled enough so the address
>> increment doesn't matter.
>>
>>> There is the piece of disassemble code:
>>>    120000d90:   28c001f0        ld.d    $r16,$r15,0
>>>    120000d94:   0010c58c        add.d   $r12,$r12,$r17
>>>    120000d98:   02c021ef        addi.d  $r15,$r15,8(0x8)
>>
>> Those three instructions are independent.
>>
>>>    120000d9c:   0010b20c        add.d   $r12,$r16,$r12
>>
>> that one depends on the ld.d
>>
>>>    120000da0:   0012c191        sltu    $r17,$r12,$r16
>>
>> that depends on the add.d
>> but it could be execute after the 'bne' in parallel with the ld.d
>>
>>>    120000da4:   5fffedf2        bne     $r15,$r18,-20(0x3ffec) # 120000d90 <do_csum_64+0x90>
>>
>> If you tweak the code it is possible to get down to just
>> the addi.d and bne constraining the dependency chain.
>> (Assuming there is no delay on the read and there are an infinite
>> number of execution units.)
>> Unroll once and do:
>>         ld.d r,addr,0
>>         addi.d addr,16
>>         ld.d r,addr,-8
>>         bne addr,limit,loop_top
>> and you might get a loop that does a memory read every clock.
>>
>> So you end up worrying about how the memory read delays affect
>> the instruction pipeline.
>> The Intel x86 cpu I've got just pile up the arithmetic instructions
>> waiting for the data to be read.
>> If you get a memory read requested every clock everything else
>> follows - provided you don't try to execute too many instrcutions
>> at once.
>>
>>         David
>>
>> -
>> Registered Address Lakeside, Bramley Road, Mount Farm, Milton Keynes, MK1 1PT, UK
>> Registration No: 1397386 (Wales)

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