为什么_umul128的运行速度比mul128x64x2函数的标量代码慢? [英] Why _umul128 works slower than scalar code for mul128x64x2 function?

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本文介绍了为什么_umul128的运行速度比mul128x64x2函数的标量代码慢?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我第二次尝试实现mul128x64x2快速功能. 第一次我问这个问题而没有与_umul128进行比较MSVC版本.现在,我进行了这样的比较,结果表明_umul128函数的速度比本机标量和​​手工simd AVX 1.0代码慢.

I am second time trying to implement fast mul128x64x2 function. First time I ask the question without comparision with _umul128 MSVC version. Now I made such a comparison and the results that I got show that the _umul128 function slower then native scalar and handmade simd AVX 1.0 code.

在我的测试代码下面:

#include <iostream>
#include <chrono>

#include <intrin.h>
#include <emmintrin.h>
#include <immintrin.h>

#pragma intrinsic(_umul128)

constexpr uint32_t LOW[4] = { 4294967295u, 0u, 4294967295u, 0u };

__forceinline void multiply128x128( const uint32_t ABCD[4], const uint32_t EFGH[4], uint32_t OUT[2][4] ) noexcept
{
    __m128i L  = _mm_lddqu_si128( reinterpret_cast< __m128i const* >( LOW ) );
    __m128i IN = _mm_lddqu_si128( reinterpret_cast< __m128i const* >( EFGH ) );

    __m128i A  = _mm_set1_epi32( ABCD[0] );
    __m128i B  = _mm_set1_epi32( ABCD[1] );
    __m128i C  = _mm_set1_epi32( ABCD[2] );
    __m128i D  = _mm_set1_epi32( ABCD[3] );

    __m128i ED = _mm_mul_epu32( IN, D );
    __m128i EC = _mm_mul_epu32( IN, C );
    __m128i EB = _mm_mul_epu32( IN, B );
    __m128i EA = _mm_mul_epu32( IN, A );

    IN = _mm_srli_epi64( IN, 32 );

    __m128i FD = _mm_mul_epu32( IN, D );
    __m128i FC = _mm_mul_epu32( IN, C );
    __m128i FB = _mm_mul_epu32( IN, B );
    __m128i FA = _mm_mul_epu32( IN, A );

    __m128i FD_H = _mm_srli_epi64( FD, 32 );
    __m128i FD_L = _mm_and_si128 ( L, FD );

    __m128i FC_H = _mm_srli_epi64( FC, 32 );
    __m128i FC_L = _mm_and_si128 ( L, FC );

    __m128i FB_H = _mm_srli_epi64( FB, 32 );
    __m128i FB_L = _mm_and_si128 ( L, FB );

    __m128i FA_H = _mm_srli_epi64( FA, 32 );
    __m128i FA_L = _mm_and_si128 ( L, FA );

    __m128i ED_H = _mm_srli_epi64( ED, 32 );
    __m128i ED_L = _mm_and_si128 ( L, ED );

    __m128i EC_H = _mm_srli_epi64( EC, 32 );
    __m128i EC_L = _mm_and_si128 ( L, EC );

    __m128i EB_H = _mm_srli_epi64( EB, 32 );
    __m128i EB_L = _mm_and_si128 ( L, EB );

    __m128i EA_H = _mm_srli_epi64( EA, 32 );
    __m128i EA_L = _mm_and_si128 ( L, EA );

    __m128i SUM_FC_L_FD_H = _mm_add_epi64( FC_L, FD_H );
    __m128i SUM_FB_L_FC_H = _mm_add_epi64( FB_L, FC_H );
    __m128i SUM_FA_L_FB_H = _mm_add_epi64( FA_L, FB_H );

    __m128i SUM_EC_L_ED_H = _mm_add_epi64( EC_L, ED_H );
    __m128i SUM_EB_L_EC_H = _mm_add_epi64( EB_L, EC_H );
    __m128i SUM_EA_L_EB_H = _mm_add_epi64( EA_L, EB_H );

    __m128i SUM_FC_L_FD_H_ED_L         = _mm_add_epi64( SUM_FC_L_FD_H, ED_L );
    __m128i SUM_FB_L_FC_H_EC_L_ED_H    = _mm_add_epi64( SUM_FB_L_FC_H, SUM_EC_L_ED_H );
    __m128i SUM_FA_L_FB_H_EB_L_EC_H    = _mm_add_epi64( SUM_FA_L_FB_H, SUM_EB_L_EC_H );
    __m128i SUM_FA_H_EA_L_EB_H         = _mm_add_epi64( FA_H, SUM_EA_L_EB_H );

    __m128i SUM_FC_L_FD_H_ED_L_L       = _mm_srli_epi64( SUM_FC_L_FD_H_ED_L, 32 );
            SUM_FC_L_FD_H_ED_L_L       = _mm_add_epi64 ( SUM_FC_L_FD_H_ED_L_L, SUM_FB_L_FC_H_EC_L_ED_H );

    __m128i SUM_FC_L_FD_H_ED_L_L_L     = _mm_srli_epi64( SUM_FC_L_FD_H_ED_L_L, 32 );
            SUM_FC_L_FD_H_ED_L_L_L     = _mm_add_epi64 ( SUM_FC_L_FD_H_ED_L_L_L, SUM_FA_L_FB_H_EB_L_EC_H );

    __m128i SUM_FC_L_FD_H_ED_L_L_L_L   = _mm_srli_epi64( SUM_FC_L_FD_H_ED_L_L_L, 32 );
            SUM_FC_L_FD_H_ED_L_L_L_L   = _mm_add_epi64 ( SUM_FC_L_FD_H_ED_L_L_L_L, SUM_FA_H_EA_L_EB_H );

    __m128i SUM_FC_L_FD_H_ED_L_L_L_L_L = _mm_srli_epi64( SUM_FC_L_FD_H_ED_L_L_L_L, 32 );
            SUM_FC_L_FD_H_ED_L_L_L_L_L = _mm_add_epi64 ( SUM_FC_L_FD_H_ED_L_L_L_L_L, EA_H );

    OUT[0][0] = SUM_FC_L_FD_H_ED_L_L_L_L_L.m128i_u32[0];
    OUT[0][1] = SUM_FC_L_FD_H_ED_L_L_L_L.m128i_u32[0];
    OUT[0][2] = SUM_FC_L_FD_H_ED_L_L_L.m128i_u32[0];
    OUT[0][3] = SUM_FC_L_FD_H_ED_L_L.m128i_u32[0];

    OUT[1][0] = SUM_FC_L_FD_H_ED_L_L_L_L_L.m128i_u32[2];
    OUT[1][1] = SUM_FC_L_FD_H_ED_L_L_L_L.m128i_u32[2];
    OUT[1][2] = SUM_FC_L_FD_H_ED_L_L_L.m128i_u32[2];
    OUT[1][3] = SUM_FC_L_FD_H_ED_L_L.m128i_u32[2];
}


__forceinline void multiply128x128_1( const uint32_t ABCD[4], const uint32_t EFGH[4], uint32_t OUT[2][4] ) noexcept
{
    uint64_t ED = static_cast<uint64_t>( ABCD[3] ) * static_cast<uint64_t>( EFGH[0] );
    uint64_t EC = static_cast<uint64_t>( ABCD[2] ) * static_cast<uint64_t>( EFGH[0] );
    uint64_t EB = static_cast<uint64_t>( ABCD[1] ) * static_cast<uint64_t>( EFGH[0] );
    uint64_t EA = static_cast<uint64_t>( ABCD[0] ) * static_cast<uint64_t>( EFGH[0] );

    uint64_t FD = static_cast<uint64_t>( ABCD[3] ) * static_cast<uint64_t>( EFGH[1] );
    uint64_t FC = static_cast<uint64_t>( ABCD[2] ) * static_cast<uint64_t>( EFGH[1] );
    uint64_t FB = static_cast<uint64_t>( ABCD[1] ) * static_cast<uint64_t>( EFGH[1] );
    uint64_t FA = static_cast<uint64_t>( ABCD[0] ) * static_cast<uint64_t>( EFGH[1] );

    uint64_t GD = static_cast<uint64_t>( ABCD[3] ) * static_cast<uint64_t>( EFGH[2] );
    uint64_t GC = static_cast<uint64_t>( ABCD[2] ) * static_cast<uint64_t>( EFGH[2] );
    uint64_t GB = static_cast<uint64_t>( ABCD[1] ) * static_cast<uint64_t>( EFGH[2] );
    uint64_t GA = static_cast<uint64_t>( ABCD[0] ) * static_cast<uint64_t>( EFGH[2] );

    uint64_t HD = static_cast<uint64_t>( ABCD[3] ) * static_cast<uint64_t>( EFGH[3] );
    uint64_t HC = static_cast<uint64_t>( ABCD[2] ) * static_cast<uint64_t>( EFGH[3] );
    uint64_t HB = static_cast<uint64_t>( ABCD[1] ) * static_cast<uint64_t>( EFGH[3] );
    uint64_t HA = static_cast<uint64_t>( ABCD[0] ) * static_cast<uint64_t>( EFGH[3] );

    uint64_t SUM_FC_L_FD_H = ( FC & 0xFFFFFFFF ) + ( FD >> 32u );
    uint64_t SUM_FB_L_FC_H = ( FB & 0xFFFFFFFF ) + ( FC >> 32u );
    uint64_t SUM_FA_L_FB_H = ( FA & 0xFFFFFFFF ) + ( FB >> 32u );

    uint64_t SUM_EC_L_ED_H = ( EC & 0xFFFFFFFF ) + ( ED >> 32u );
    uint64_t SUM_EB_L_EC_H = ( EB & 0xFFFFFFFF ) + ( EC >> 32u );
    uint64_t SUM_EA_L_EB_H = ( EA & 0xFFFFFFFF ) + ( EB >> 32u );

    uint64_t SUM_HC_L_HD_H = ( HC & 0xFFFFFFFF ) + ( HD >> 32u );
    uint64_t SUM_HB_L_HC_H = ( HB & 0xFFFFFFFF ) + ( HC >> 32u );
    uint64_t SUM_HA_L_HB_H = ( HA & 0xFFFFFFFF ) + ( HB >> 32u );

    uint64_t SUM_GC_L_GD_H = ( GC & 0xFFFFFFFF ) + ( GD >> 32u );
    uint64_t SUM_GB_L_GC_H = ( GB & 0xFFFFFFFF ) + ( GC >> 32u );
    uint64_t SUM_GA_L_GB_H = ( GA & 0xFFFFFFFF ) + ( GB >> 32u );

    uint64_t SUM_FC_L_FD_H_ED_L         = SUM_FC_L_FD_H + ( ED & 0xFFFFFFFF );
    uint64_t SUM_FB_L_FC_H_EC_L_ED_H    = SUM_FB_L_FC_H + SUM_EC_L_ED_H;
    uint64_t SUM_FA_L_FB_H_EB_L_EC_H    = SUM_FA_L_FB_H + SUM_EB_L_EC_H;
    uint64_t SUM_FA_H_EA_L_EB_H         = SUM_EA_L_EB_H + ( FA >> 32u );

    uint64_t SUM_FC_L_FD_H_ED_L_L       = ( SUM_FC_L_FD_H_ED_L       >> 32u ) + SUM_FB_L_FC_H_EC_L_ED_H;
    uint64_t SUM_FC_L_FD_H_ED_L_L_L     = ( SUM_FC_L_FD_H_ED_L_L     >> 32u ) + SUM_FA_L_FB_H_EB_L_EC_H;
    uint64_t SUM_FC_L_FD_H_ED_L_L_L_L   = ( SUM_FC_L_FD_H_ED_L_L_L   >> 32u ) + SUM_FA_H_EA_L_EB_H;
    uint64_t SUM_FC_L_FD_H_ED_L_L_L_L_L = ( SUM_FC_L_FD_H_ED_L_L_L_L >> 32u ) + ( EA >> 32u );

    uint64_t SUM_HC_L_HD_H_GD_L         = SUM_HC_L_HD_H + ( GD & 0xFFFFFFFF );
    uint64_t SUM_HB_L_HC_H_GC_L_GD_H    = SUM_HB_L_HC_H + SUM_GC_L_GD_H;
    uint64_t SUM_HA_L_HB_H_GB_L_GC_H    = SUM_HA_L_HB_H + SUM_GB_L_GC_H;
    uint64_t SUM_HA_H_GA_L_GB_H         = SUM_GA_L_GB_H + ( HA >> 32u );

    uint64_t SUM_HC_L_HD_H_GD_L_L       = ( SUM_HC_L_HD_H_GD_L       >> 32u ) + SUM_HB_L_HC_H_GC_L_GD_H;
    uint64_t SUM_HC_L_HD_H_GD_L_L_L     = ( SUM_HC_L_HD_H_GD_L_L     >> 32u ) + SUM_HA_L_HB_H_GB_L_GC_H;
    uint64_t SUM_HC_L_HD_H_GD_L_L_L_L   = ( SUM_HC_L_HD_H_GD_L_L_L   >> 32u ) + SUM_HA_H_GA_L_GB_H;
    uint64_t SUM_HC_L_HD_H_GD_L_L_L_L_L = ( SUM_HC_L_HD_H_GD_L_L_L_L >> 32u ) + ( GA >> 32u );

    OUT[0][0] = SUM_FC_L_FD_H_ED_L_L_L_L_L;
    OUT[0][1] = SUM_FC_L_FD_H_ED_L_L_L_L;
    OUT[0][2] = SUM_FC_L_FD_H_ED_L_L_L;
    OUT[0][3] = SUM_FC_L_FD_H_ED_L_L;

    OUT[1][0] = SUM_HC_L_HD_H_GD_L_L_L_L_L;
    OUT[1][1] = SUM_HC_L_HD_H_GD_L_L_L_L;
    OUT[1][2] = SUM_HC_L_HD_H_GD_L_L_L;
    OUT[1][3] = SUM_HC_L_HD_H_GD_L_L;
}


__forceinline void mulShift( const uint64_t* const m, const uint64_t* const mul , uint32_t OUT[2][4]) noexcept
{
    uint64_t B0[2];
    uint64_t B2[2];

    {
        B0[0] = _umul128( m[1], mul[0], &B0[1] );
        B2[0] = _umul128( m[0], mul[0], &B2[1] );

        uint64_t S = B0[1] + B2[0];

        OUT[0][2] = S >> 32;
        OUT[0][3] = S & 0xFFFFFFFF;

        uint64_t M = B2[1] + ( S < B2[0] );

        OUT[0][1] = M & 0xFFFFFFFF;
        OUT[0][0] = M >> 32;
    }

    {
        B0[0] = _umul128( m[1], mul[1], &B0[1] );
        B2[0] = _umul128( m[0], mul[1], &B2[1] );

        uint64_t S = B0[1] + B2[0];

        OUT[1][2] = S >> 32;
        OUT[1][3] = S & 0xFFFFFFFF;

        uint64_t M = B2[1] + ( S < B2[0] );

        OUT[1][1] = M & 0xFFFFFFFF;
        OUT[1][0] = M >> 32;
    }
}


constexpr uint32_t N = 1 << 28;

int main()
{
    uint32_t OUT[2][4];

    uint32_t ABCD[4] = { 4294967295u, 4294967295u, 4294967295u, 4294967295u };
    uint32_t EFGH[4] = { 4294967295u, 4294967295u, 4294967295u, 4294967295u };

    multiply128x128_1( ABCD, EFGH, OUT );

    uint64_t S_1 = 0u;
    uint64_t S_2 = 0u;
    uint64_t S_3 = 0u;

    auto start_1 = std::chrono::high_resolution_clock::now();

    for ( uint32_t i = 0; i < N; ++i )
    {
        EFGH[0] = i;
        EFGH[1] = i;
        EFGH[2] = i + 1;
        EFGH[3] = i + 1;

        ABCD[0] = i;
        ABCD[1] = i;
        ABCD[2] = i + 1;
        ABCD[3] = i + 1;

        multiply128x128( ABCD, EFGH, OUT );

        S_1 += OUT[0][0] + OUT[0][1] + OUT[0][2] + OUT[0][3];
        S_1 += OUT[1][0] + OUT[1][1] + OUT[1][2] + OUT[1][3];
    }

    auto stop_1 = std::chrono::high_resolution_clock::now();
    std::cout << "Test A: " << std::chrono::duration_cast<std::chrono::milliseconds>( stop_1 - start_1 ).count() << '\n';

    auto start_2 = std::chrono::high_resolution_clock::now();


    for ( uint32_t i = 0; i < N; ++i )
    {
        EFGH[0] = i;
        EFGH[1] = i;
        EFGH[2] = i + 1;
        EFGH[3] = i + 1;

        ABCD[0] = i;
        ABCD[1] = i;
        ABCD[2] = i + 1;
        ABCD[3] = i + 1;

       mulShift( reinterpret_cast<const uint64_t*>( ABCD ), reinterpret_cast<const uint64_t*>( EFGH ), OUT );
       S_2 += OUT[0][0] + OUT[0][1] + OUT[0][2] + OUT[0][3];
       S_2 += OUT[1][0] + OUT[1][1] + OUT[1][2] + OUT[1][3];
    }

    auto stop_2 = std::chrono::high_resolution_clock::now();
    std::cout << "Test B: " << std::chrono::duration_cast<std::chrono::milliseconds>( stop_2 - start_2 ).count() << '\n';


    auto start_3 = std::chrono::high_resolution_clock::now();

    for ( uint32_t i = 0; i < N; ++i )
    {
        EFGH[0] = i;
        EFGH[1] = i;
        EFGH[2] = i + 1;
        EFGH[3] = i + 1;

        ABCD[0] = i;
        ABCD[1] = i;
        ABCD[2] = i + 1;
        ABCD[3] = i + 1;

        multiply128x128_1( ABCD, EFGH, OUT );

        S_3 += OUT[0][0] + OUT[0][1] + OUT[0][2] + OUT[0][3];
        S_3 += OUT[1][0] + OUT[1][1] + OUT[1][2] + OUT[1][3];
    }

    auto stop_3 = std::chrono::high_resolution_clock::now();
    std::cout << "Test C: " << std::chrono::duration_cast<std::chrono::milliseconds>( stop_3 - start_3 ).count() << '\n';

    std::cout << S_1 << " " << S_2 << " " <<  S_3 << '\n';
}

_umul128为什么这么慢?也许我在上面的测试代码中犯了一些错误?

Why is _umul128 so slow? Maybe i did some mistakes in my test code above?

我的结果: 测试A(模拟):4546ms. 测试B(_umul128):6637ms. 测试C(标量):2333ms.

My results: Test A (simd): 4546ms. Test B (_umul128): 6637ms. Test C (scalar): 2333ms.

在Windows 10,x64,MSVC 2019上进行了测试

Tested on Windows 10, x64, MSVC 2019

推荐答案

_umul128版本并不是真的那么慢,但是您通过弄乱商店转发摊位来讨价还价使MSVC发出可怕的asm的32位数组.

The _umul128 version isn't really that slow but you're gimping it with store-forwarding stalls by messing around with 32-bit arrays that makes MSVC emit terrible asm.

优化正在超越您的基准;纯C版本并不是真的那么快.

特别是简单的输入数据:

Especially with the simple input data:

       ABCD[0] = EFGH[0] = i;
       ABCD[1] = EFGH[1] = i;
       ABCD[2] = EFGH[2] = i + 1;
       ABCD[3] = EFGH[3] = i + 1;

像这样初始化两个输入会在内联纯C版本后为优化创造大量机会.它执行i*i 4次,并且i*(i+1) = i*i + i另外8次,并且(i+1)*(i+1) 4次. MSVC并不傻,并且注意到这一点. 这称为消除常见子表达式(CSE).

Initializing both inputs like this creates a huge amount of opportunity for optimization after inlining the pure C version. It does i*i 4 times, and i*(i+1) = i*i + i another 8 times, and also (i+1)*(i+1) 4 times. MSVC isn't dumb and notices this. This is called Common Subexpression Elimination (CSE).

如果您想了解纯C语言的实际运行速度,则需要提出一种更复杂的方法来伪造输入.也许提前生成然后在包含输入的内存上循环?从循环计数器设置输入的成本几乎等于乘积.

You'll need to come up with a more sophisticated way to fake input if you want to see how slow the pure C really is. Maybe generate ahead of time then loop over memory containing inputs? Setting up inputs from a loop counter costs almost as much as a multiply.

MSVC的asm输出确认大部分工作已针对纯C版本进行了优化.(使用MSVC 19.22 for x64的Godbolt )

   ...
$LL10@main:
        lea     r15, QWORD PTR [rax+1]
        mov     rcx, r15
        mov     r9, r15
        imul    rcx, rax               # only 3, not 16, imul instructions.
        imul    rax, rax               # (None appear later in this loop in the ... part)
        imul    r9, r15
        mov     edi, ecx
        mov     r14, rcx
        mov     r8d, eax
        shr     r14, 32                             ; 00000020H
        shr     rax, 32                             ; 00000020H
     ...
        sub     r13, 1
        jne     $LL10@main


MSVC不擅长优化内在函数,并且执行所有4条mul m64指令,而不是注意ii * i1i1进行了两次.


MSVC is bad at optimizing intrinsics and does all 4 mul m64 instructions instead of noticing that ii * i1i1 is done twice.

更重要的是, _umul128循环因存储转发停顿而受到损害,因为它实际上将数组存储到具有32位存储的内存中,然后使用64位位加载以馈送mul m64.

More importantly, the _umul128 loop is hurt by store-forwarding stalls because it actually stores your array to memory with 32-bit stores and then uses 64-bit loads to feed mul m64.

此外,以32位块形式处理输出只会使您大吃一惊,引入了额外的移位和mov操作.

Also, handling the output in 32-bit chunks just shoots yourself in the foot, introducing extra shifts and mov operations.

这并不复杂,实际上只需要3条指令mul r64imul r64, r64加上上半部的add. GCC/clang很容易发出正确的消息,并且x86-64 System V调用约定可以在寄存器中返回128位int.

This is not complicated, literally just 3 instructions, mul r64 and imul r64, r64 plus an add for the high half, is all that's needed. GCC/clang easily emit the right thing, and the x86-64 System V calling convention can return a 128-bit int in registers.

在Godbolt上: https://godbolt.org/z/DcZhSl >

On Godbolt: https://godbolt.org/z/DcZhSl

#include <stdint.h>
#ifdef __GNUC__
typedef unsigned __int128 u128;

u128 mul128x64( u128 a, uint64_t b) {
    return a * b;
}
#endif

# clang -O3 for the x86-64 System V ABI (Linux)
mul128x64(unsigned __int128, unsigned long):                         # 
    mov     rax, rdi
    imul    rsi, rdx
    mul     rdx
    add     rdx, rsi
    ret

对于MSVC,我们必须自己做,调用约定意味着结果将返回到内存中.

For MSVC we have to do that ourself, and the calling convention means the result is returned in memory.

#ifdef _MSC_VER
#include <intrin.h>

struct u128 { uint64_t u64[2]; };
u128 mul128x64( uint64_t a_lo, uint64_t a_hi, uint64_t b)
{
    uint64_t lolo_high;
    uint64_t lolo = _umul128( a_lo, b, &lolo_high );
    uint64_t lohi = a_hi * b;
    return {{lolo, lohi + lolo_high}};
}
#endif

# MSVC x64 -O2 
u128 mul128x64(unsigned __int64,unsigned __int64,unsigned __int64) PROC
    mov     rax, r9
    mul     rdx
    imul    r8, r9
    mov     QWORD PTR [rcx], rax         # store the retval into hidden pointer
    mov     rax, rcx
    add     r8, rdx
    mov     QWORD PTR [rcx+8], r8
    ret     0


您的__m128i内部函数版本不太可能会获胜.现代x86(主流Intel SnB系列,AMD Ryzen)具有mulimul的1/时钟吞吐量. (除了Ryzen,加宽i/mul r64的吞吐量为2c,但imul r64,r64的吞吐量为1/时钟.)


Your __m128i intrinsics version is unlikely to be a win. Modern x86 (mainstream Intel SnB-family, AMD Ryzen) has 1/clock throughput for mul and imul. (Except Ryzen where widening i/mul r64 has 2c throughput, but still 1/clock for imul r64,r64.)

因此,如果您使用C编译成这样的asm,在Sandybridge系列上进行64 x 128位乘法的总吞吐量为每2个周期之一(瓶颈在端口1上).

So overall throughput for a 64 x 128-bit multiply on Sandybridge-family is one per 2 cycles (bottlenecked on port 1), if you implement in C that compiles to asm like this.

鉴于您需要四个以上的pmuludq指令来实现乘法,所以AVX1是非入门级的. (Skylake的pmuludq吞吐量为0.5c.Sandybridge的吞吐量为1c,因此您必须以平均每2倍的pmuludq insns来完成工作才能与标量竞争.而且,这无需考虑所有的偏移/随机播放/添加需要做的工作.

Given that you need more than 4 pmuludq instructions to implement a multiply, AVX1 is a non-starter. (Skylake has 0.5c throughput for pmuludq. Sandybridge has 1c throughput so you'd need to get the job done in 2 pmuludq insns per multiply (on average) to compete with scalar. And that's without considering all the shift / shuffle / add work that needs doing.

可能值得考虑Bulldozer系列,其中64位标量乘法为4c吞吐量,而pmuludq为1c. ( https://agner.org/optimize/)每个周期产生128个乘积位(两个32x32 => 64位产品)要比每4个周期产生128个产品位好,如果您可以在不占用过多额外周期的情况下进行转换和添加.

Possibly worth considering on Bulldozer-family where 64-bit scalar multiply is 4c throughput but pmuludq is 1c. (https://agner.org/optimize/) Producing 128 product bits per cycle (two 32x32 => 64-bit products) is better than producing 128 product bits per 4 cycles, if you can get them shifted and added without eating up too many extra cycles.

同样,MSVC不利于通过内部常数进行恒定传播或CSE优化,因此您的内在版本不会从中受益.

Again, MSVC is bad at constant-propagation or CSE optimization through intrinsincs, so your intrinsics version doesn't benefit from anything.

您的测试代码还使用标量整数循环变量中的_mm_set1_epi32( ),需要vmovdvpshufd指令.

Your test code also uses _mm_set1_epi32( ) from scalar integer loop variables, requiring vmovd and vpshufd instructions.

您将在这些数组上为lddqu内部函数进行标量存储/向量重载,因此再次有存储转发停顿.

And you get scalar store / vector reload for the lddqu intrinsics on those arrays, so again you have store-forwarding stalls.

使用SSE2或AVX1实现此目的的唯一希望是,您的数据来自内存,而不是寄存器.或者,如果您可以将数据长时间保存在向量寄存器中,而不要不断地来回移动它.特别是在int<-> SIMD具有高延迟的推土机家族中.

The only hope for this being good with SSE2 or AVX1 is if your data comes from memory, not registers. Or if you can keep your data in vector registers for a long time, not constantly moving it back and forth. Especially on Bulldozer-family where int <-> SIMD has high latency.

这篇关于为什么_umul128的运行速度比mul128x64x2函数的标量代码慢?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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