Question

我在代码中有以下关键位置：我需要从大约1＆＃39; 000＆000,000次的64字节数组中查找。

最小代码：

#include <iostream>
#include <stdint.h>
#include <random>
#include <chrono>
#include <ctime>
#define TYPE uint8_t
#define n_lookup 64

int main(){
    const int n_indices = 1000000;
    TYPE lookup[n_lookup];
    TYPE indices[n_indices];
    TYPE result[n_indices];

    // preparations
    std::default_random_engine generator;
    std::uniform_int_distribution<int> distribution(0, n_lookup);
    for (int i=0; i < n_indices; i++) indices[i] = distribution(generator);
    for (int i=0; i < n_lookup; i++) lookup[i] = distribution(generator);

    std::chrono::time_point<std::chrono::system_clock> start = std::chrono::system_clock::now();
    // main loop:
    for (int i=0; i < n_indices; i++) {
        result[i] = lookup[indices[i]];
    }
    std::chrono::time_point<std::chrono::system_clock> end = std::chrono::system_clock::now();

    std::chrono::duration<double> elapsed_seconds = end - start;
    std::cout << "computation took " << elapsed_seconds.count() * 1e9 / n_indices  <<  " ns per element"<< std::endl;
    // printing random numbers to avoid code elimination
    std::cout << result[12] << result[45]; 
    return 0;
}

使用g++ lookup.cpp -std=gnu++11 -O3 -funroll-loops进行编译后，在现代CPU上，每个元素的值不到1ns。

我需要这个操作加快2-3倍（没有线程）。我怎么能这样做？

P.S。我也在调查AVX512（512位正是查找表的大小！）指令集，但它缺少8位聚集操作！

Answer 1

indices和result向量位于内存中的不同位置，但同时访问。它导致缓存未命中。我建议你在一个向量中合并结果和索引。这是代码：

#include <iostream>
#include <stdint.h>
#include <random>
#include <chrono>
#include <ctime>
#define TYPE uint8_t
#define n_lookup 64

int main(){
    const int n_indices = 2000000;
    TYPE lookup[n_lookup];
    // Merge indices and result
    // If i is index, then i+1 is result
    TYPE ind_res[n_indices];

    // preparations
    std::default_random_engine generator;
    std::uniform_int_distribution<int> distribution(0, n_lookup);
    for (int i=0; i < n_indices; i += 2) ind_res[i] = distribution(generator);
    for (int i=0; i < n_lookup; i++) lookup[i] = distribution(generator);

    std::chrono::time_point<std::chrono::system_clock> start = std::chrono::system_clock::now();
    // main loop:
    for (int i=0; i < n_indices; i += 2) {
        ind_res[i+1] = lookup[ind_res[i]]; // more dense access here, no cache-miss
    }
    std::chrono::time_point<std::chrono::system_clock> end = std::chrono::system_clock::now();

    std::chrono::duration<double> elapsed_seconds = end - start;
    std::cout << "computation took " << elapsed_seconds.count() * 1e9 / n_indices  <<  " ns per element"<< std::endl;
    // printing random numbers to avoid code elimination
    std::cout << ind_res[24] << ind_res[90]; 
    return 0;
}

我的测试显示此代码运行得更快。

Answer 2

with -march = native这是你的循环编译成的：

        movq    %rax, %rbx
        xorl    %eax, %eax
.L145:
        movzbl  128(%rsp,%rax), %edx
        movzbl  64(%rsp,%rdx), %edx
        movb    %dl, 1000128(%rsp,%rax)
        addq    $1, %rax
        cmpq    $1000000, %rax
        jne     .L145

如果没有并行化，我很难看到如何更快。

通过将TYPE更改为int32_t，它会被矢量化：

    vpcmpeqd        %ymm2, %ymm2, %ymm2
    movq    %rax, %rbx
    xorl    %eax, %eax
.L145:
        vmovdqa -8000048(%rbp,%rax), %ymm1
        vmovdqa %ymm2, %ymm3
        vpgatherdd      %ymm3, -8000304(%rbp,%ymm1,4), %ymm0
        vmovdqa %ymm0, -4000048(%rbp,%rax)
        addq    $32, %rax
        cmpq    $4000000, %rax
        jne     .L145
        vzeroupper

可能有帮助吗？

Answer 3

首先，有一个错误，分布（0,64）产生数字0到64,64不能适合数组。

您可以通过一次查找两个值来加快查找速度：

#include <iostream>
#include <stdint.h>
#include <random>
#include <chrono>
#include <ctime>
#define TYPE uint8_t
#define TYPE2 uint16_t
#define n_lookup 64

void tst() {
    const int n_indices = 1000000;// has to be multiple of 2
    TYPE lookup[n_lookup];
    TYPE indices[n_indices];
    TYPE result[n_indices];
    TYPE2 lookup2[n_lookup * 256];

    // preparations
    std::default_random_engine generator;
    std::uniform_int_distribution<int> distribution(0, n_lookup-1);
    for (int i = 0; i < n_indices; i++) indices[i] = distribution(generator);
    for (int i = 0; i < n_lookup; i++) lookup[i] = distribution(generator);

    for (int i = 0; i < n_lookup; ++i) {
        for (int j = 0; j < n_lookup; ++j) {
            lookup2[(i << 8) | j] = (lookup[i] << 8) | lookup[j];
        }
    }

    std::chrono::time_point<std::chrono::system_clock> start = std::chrono::system_clock::now();

    TYPE2* indices2 = (TYPE2*)indices;
    TYPE2* result2 = (TYPE2*)result;

    // main loop:
    for (int i = 0; i < n_indices / 2; ++i) {
        *result2++ = lookup2[*indices2++];
    }

    std::chrono::time_point<std::chrono::system_clock> end = std::chrono::system_clock::now();

    for (int i = 0; i < n_indices; i++) {
        if (result[i] != lookup[indices[i]]) {
            std::cout << "!!!!!!!!!!!!!ERROR!!!!!!!!!!!!!";
        }
    }

    std::chrono::duration<double> elapsed_seconds = end - start;
    std::cout << "computation took " << elapsed_seconds.count() * 1e9 / n_indices << " ns per element" << std::endl;
    // printing random numbers to avoid code elimination
    std::cout << result[12] << result[45];
}

int main() {
    tst();
    std::cin.get();
    return 0;
}

Answer 4

您的代码已经非常快了。然而（在我的系统上）当你改变

时，执行速度大约快4.858％

const int n_indices = 1048576; // 2^10

到

FBSDKLoginKit

这并不多，但它确实存在。

从64字节数组中查找字节的最快方法

4 个答案: