我正在尝试找到一种方法来击败硬件预取器以检测流模式并以随机顺序访问 4KB 数据 这样它就不会被 H/w 预取器检测到和预取。

最初我想以随机模式访问所有偶数索引数据，因为 H/w 预取器预取下一个缓存行 总是(所以当我访问偶数索引时，下一个奇数索引数据已经被预取)。

我编写代码以随机模式访问所有偶数索引数据，但结果表明预取器检测到该模式 (不知道如何？没有固定步幅，都是随机步幅)

我正在调查发生这种情况的原因，然后我在 Intel 中找到了这篇文章； https://software.intel.com/en-us/forums/topic/473493

根据 John D. McCalpin 博士的说法，“Bandwidth 博士，

In section 2.2.5.4 of "Intel 64 and IA-32 Architectures Optimization Reference Manual" (document 248966-028, July 2013),it states that,

streamer prefetcher "[d]etects and maintains up to 32 streams of data accesses. For each 4K byte page, you can maintain one forward and one backward stream can be maintained.

This implies that the L2 hardware prefetcher tracks the 16 4KiB pages most recently accessed and remembers enough of the access patterns for those pages to track one forward stream and one backward stream. So to defeat the L2 streamer prefetcher with "random" fetches, simply ensure that you access more than 15 other 4 KiB pages before you make a second reference to a previously referenced page. So a "random" sequences of fetches might be composed of a random permutation of more than 16 4 KiB page numbers with a random offset within each page. (I typically use at least 32 pages in my permutation list.)

因此这意味着在访问相同 4KB 页面的两个不同随机索引之间，我们需要访问至少 16 个 4KB 页面才能击败硬件预取器。

我已经实现了 John D. McCalpin 建议的概念，但是结果再次显示硬件预取器没有被击败。它能够检测某些模式并预取数据(参见示例输出)。我访问的页面数量从 20-40 个 4KB 页面不等，但结果没有改善/变化。

这是我的代码:

#define _GNU_SOURCE             /* See feature_test_macros(7) */
#include <stdlib.h>
#include <stdio.h>
#include <unistd.h>
#include <sched.h>

#ifndef _POSIX_THREAD_PROCESS_SHARED
#error This system does not support process shared mutex
#endif

#define MAX_COUNT 3000
#define INDEX (40*1024) // size of DUMMY 40 4KB pages

inline void clflush(volatile void *p)
{
    asm volatile ("clflush (%0)" :: "r"(p));
}

unsigned long probe(char *adrs) {
  volatile unsigned long time;
  asm __volatile__ (
    " mfence              \n"
    " lfence              \n"
    " rdtsc               \n"
    " lfence              \n"
    " movl %%eax, %%esi \n"
    " movl (%1), %%eax     \n"
    " lfence              \n"
    " rdtsc               \n"
    " subl %%esi, %%eax \n"
    " clflush 0(%1)       \n"
    : "=a" (time)
    : "c" (adrs)
    : "%esi", "%edx");
  return time;
}

void shuffle(int *arr, size_t n)
{
    if (n > 1) 
    {
        size_t i;
        srand(time(NULL));
        for (i = 0; i < n - 1; i++) 
        {
          size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
          int t = arr[j];
          arr[j] = arr[i];
          arr[i] = t;
        }
    }
}


static const int DATA[1024]={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160,161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176,177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199,200,201,202,203,204,205,206,207,208,209,210,211,212,213,214,215,216,217,218,219,220,221,222,223,224,225,226,227,228,229,230,231,232,233,234,235,236,237,238,239,240,241,242,243,244,245,246,247,248,249,250,251,252,253,254,255,256,257,258,259,260,261,262,263,264,265,266,267,268,269,270,271,272,273,274,275,276,277,278,279,280,281,282,283,284,285,286,287,288,289,290,291,292,293,294,295,296,297,298,299,300,301,302,303,304,305,306,307,308,309,310,311,312,313,314,315,316,317,318,319,320,321,322,323,324,325,326,327,328,329,330,331,332,333,334,335,336,337,338,339,340,341,342,343,344,345,346,347,348,349,350,351,352,353,354,355,356,357,358,359,360,361,362,363,364,365,366,367,368,369,370,371,372,373,374,375,376,377,378,379,380,381,382,383,384,385,386,387,388,389,390,391,392,393,394,395,396,397,398,399,400,401,402,403,404,405,406,407,408,409,410,411,412,413,414,415,416,417,418,419,420,421,422,423,424,425,426,427,428,429,430,431,432,433,434,435,436,437,438,439,440,441,442,443,444,445,446,447,448,449,450,451,452,453,454,455,456,457,458,459,460,461,462,463,464,465,466,467,468,469,470,471,472,473,474,475,476,477,478,479,480,481,482,483,484,485,486,487,488,489,490,491,492,493,494,495,496,497,498,499,500,501,502,503,504,505,506,507,508,509,510,511,512,513,514,515,516,517,518,519,520,521,522,523,524,525,526,527,528,529,530,531,532,533,534,535,536,537,538,539,540,541,542,543,544,545,546,547,548,549,550,551,552,553,554,555,556,557,558,559,560,561,562,563,564,565,566,567,568,569,570,571,572,573,574,575,576,577,578,579,580,581,582,583,584,585,586,587,588,589,590,591,592,593,594,595,596,597,598,599,600,601,602,603,604,605,606,607,608,609,610,611,612,613,614,615,616,617,618,619,620,621,622,623,624,625,626,627,628,629,630,631,632,633,634,635,636,637,638,639,640,641,642,643,644,645,646,647,648,649,650,651,652,653,654,655,656,657,658,659,660,661,662,663,664,665,666,667,668,669,670,671,672,673,674,675,676,677,678,679,680,681,682,683,684,685,686,687,688,689,690,691,692,693,694,695,696,697,698,699,700,701,702,703,704,705,706,707,708,709,710,711,712,713,714,715,716,717,718,719,720,721,722,723,724,725,726,727,728,729,730,731,732,733,734,735,736,737,738,739,740,741,742,743,744,745,746,747,748,749,750,751,752,753,754,755,756,757,758,759,760,761,762,763,764,765,766,767,768,769,770,771,772,773,774,775,776,777,778,779,780,781,782,783,784,785,786,787,788,789,790,791,792,793,794,795,796,797,798,799,800,801,802,803,804,805,806,807,808,809,810,811,812,813,814,815,816,817,818,819,820,821,822,823,824,825,826,827,828,829,830,831,832,833,834,835,836,837,838,839,840,841,842,843,844,845,846,847,848,849,850,851,852,853,854,855,856,857,858,859,860,861,862,863,864,865,866,867,868,869,870,871,872,873,874,875,876,877,878,879,880,881,882,883,884,885,886,887,888,889,890,891,892,893,894,895,896,897,898,899,900,901,902,903,904,905,906,907,908,909,910,911,912,913,914,915,916,917,918,919,920,921,922,923,924,925,926,927,928,929,930,931,932,933,934,935,936,937,938,939,940,941,942,943,944,945,946,947,948,949,950,951,952,953,954,955,956,957,958,959,960,961,962,963,964,965,966,967,968,969,970,971,972,973,974,975,976,977,978,979,980,981,982,983,984,985,986,987,988,989,990,991,992,993,994,995,996,997,998,999,1000,1001,1002,1003,1004,1005,1006,1007,1008,1009,1010,1011,1012,1013,1014,1015,1016,1017,1018,1019,1020,1021,1022,1023};

int main(int argc, char *argv[])
{

    int counter=0,k=0;
    unsigned long Access_Time[MAX_COUNT][64]={0};   
    int DUMMY[INDEX];// dummy array of 40 * 4KB ;  

    //Initialize
    for(k=0;k<INDEX;k++)
        DUMMY[k]=k;

    //access it to check segmentation fault is happening or not
    for(k=0;k<INDEX;k++)
        DUMMY[k]+=k;

    // even index in random order
    int index[32]={4,8,16,32,54,34,62,50,26,52,30,60,46,18,36,58,42,10,20,40,6,12,24,48,22,44,14,28,56,38,2,0};

    int TOTAL_RANDOM_PAGE=40;

    int i,PAGE[TOTAL_RANDOM_PAGE]; // PAGE will contain page no of 40 pages which will be accessed in random order to defeat prefetcher
        for (i=0; i<TOTAL_RANDOM_PAGE; i++)
    {
            PAGE[i] = i;
        }

    shuffle(PAGE, TOTAL_RANDOM_PAGE); // PAGE now have page no in random order

    FILE *fp2;
    int s,s1;
    int random_index=0,sum=0;

    const int *p0=&DATA[0];
    for (s=0;s<64;s++)
    {
        clflush((void *)(p0+s*16));
    }

    while(counter<MAX_COUNT)
    {               
        // Find Access time for Even Index
        for (s=0;s<32;s++)
        {

            // Access a random index
                Access_Time[counter][index[s]]=probe((char *)(p0+16*index[s]));

            //Now, access 40 other indexes belong to other 40 4KB page      
            shuffle(PAGE, TOTAL_RANDOM_PAGE); // random orderpage
            for(random_index=0;random_index<TOTAL_RANDOM_PAGE;random_index++)
            {
            DUMMY[1024*PAGE[random_index]+16*PAGE[random_index]]=2*DUMMY[1024*PAGE[random_index]+16*PAGE[random_index]];
            }

        }// end of for loop     

        // Flush all DATA from cache        
        for (s1=0;s1<64;s1++)
        {
            clflush((void *)(p0+s1*16));
        }
     counter++;

    }// end of while loop

    fp2=fopen("All_access_time.txt","a");

    int index4;
    for(counter=0;counter<MAX_COUNT;counter++)
    {
        for (index4=0;index4<64;index4++)
        {
            if(Access_Time[counter][index4]>0 && Access_Time[counter][index4]<200)
            fprintf(fp2,"%d,%d,%lu\n",counter,index4,Access_Time[counter][index4]);             
        }
    }

return 1;
}

另一个有趣的观察是，预取的随机索引的访问时间大约为 35-70 ticks。 (查看样本输出)

在我的系统中，L1 访问时间 36-44 ticks，L2 访问时间 50-70 ticks，L3 访问时间 = 90-120 ticks。

在 Intel(R) Core(TM) i3-2100 CPU @ 3.10GHz 和 Intel(R) Core(TM) i7-3770 CPU @ 3.40GHz 上进行了实验，但结果相似。

系统的一些内部细节，

L1-D = 32KB,  ways_of_associative=8
L1-I = 32KB,  ways_of_associative=8
L2   = 256KB, ways_of_associative=8
L3   = 3072KB (Core-i3), ways_of_associative=12
L3   = 8192KB (Core-i7), ways_of_associative=16
Cache line size=64Bytes

你能帮我理解为什么硬件预取器能够检测到我的随机模式吗？我哪里出错了？

如何进行编码，以便我可以击败预取器和硬件预取器无法预取我的数据？

注意:在 gcc 中使用 -O0 选项进行编译时，我禁用了软件预取器优化。

示例输出:

(counter,index,access_time)
30,8,56
30,18,72
30,20,52
30,28,72
30,34,72
30,36,72
30,38,72
30,40,72
30,42,72
31,8,52
31,18,56
31,20,52
31,28,72
31,34,52
31,36,72
31,38,56
31,40,72
31,42,52
31,60,56
32,8,52
32,18,72
32,20,52
32,28,52
32,34,72
32,36,52
32,38,72
32,40,52
32,42,52
32,48,52
33,8,56
33,18,72
33,20,52
33,28,72
33,34,52
33,36,72
33,38,72
33,40,52
33,42,72
34,8,72
34,18,52
34,20,72
34,28,72
34,34,72
34,36,52
34,38,76
34,40,72
34,42,76
34,60,72

最佳答案

如果您有足够的勇气编写内核模块，您就可以为所欲为。

作为核心 CPU 的几乎所有功能，可以禁用硬件预取逻辑以进行调试。

硬件预取由型号特定寄存器 IA32_MISC_ENABLE (0x1a0) 控制。只需设置该寄存器的第 9 位，预取器就会关闭。

更多信息请查看“Intel® 64 and IA-32 Architectures 软件开发人员手册”。搜索 IA32_MISC_ENABLE 会将您带到正确的章节。

在 Linux 内核源代码中搜索相同的关键字也有一些结果。它们与预取无关，但与预取无关，但代码看起来像是一个很好的样板，因为它展示了如何从内核读取和写入 IA32_MISC_ENABLE 寄存器。

如果你这样做，仔细检查你在做什么。您不想意外禁用热监视器。它们也位于 MISC_ENABLE :-)

关于c - 如何在 Linux 中击败核心 i3/i7 中的硬件预取器，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/31930421/