简介：本文介绍如何测试多级 cache 的访存提早，以及背地蕴含的计算机原理。
CPU 的 cache 往往是分多级的金字塔模型，L1 最靠近 CPU，拜访提早最小，但 cache 的容量也最小。本文介绍如何测试多级 cache 的访存提早，以及背地蕴含的计算机原理。
图源：https://cs.brown.edu/courses/...
Cache LatencyWikichip[1] 提供了不同 CPU 型号的 cache 提早，单位个别为 cycle，通过简略的运算，转换为 ns。以 skylake 为例，CPU 各级 cache 提早的基准值为：
CPU Frequency:2654MHz (0.3768 nanosec/clock)
设计试验1. naive thinking
申请一个 buffer，buffer size 为 cache 对应的大小，第一次遍历进行预热，将数据全副加载到 cache 中。第二次遍历统计耗时，计算每次 read 的提早平均值。
代码实现 mem-lat.c 如下：
#include <sys/types.h>#include <stdlib.h>#include <stdio.h>#include <sys/mman.h>#include <sys/time.h>#include <unistd.h>#define ONE p = (char **)*p;#define FIVE ONE ONE ONE ONE ONE#define TEN FIVE FIVE#define FIFTY TEN TEN TEN TEN TEN#define HUNDRED FIFTY FIFTYstatic void usage(){ printf("Usage: ./mem-lat -b xxx -n xxx -s xxx\n"); printf(" -b buffer size in KB\n"); printf(" -n number of read\n\n"); printf(" -s stride skipped before the next access\n\n"); printf("Please don't use non-decimal based number\n");}int main(int argc, char* argv[]){ unsigned long i, j, size, tmp; unsigned long memsize = 0x800000; /* 1/4 LLC size of skylake, 1/5 of broadwell */ unsigned long count = 1048576; /* memsize / 64 * 8 */ unsigned int stride = 64; /* skipped amount of memory before the next access */ unsigned long sec, usec; struct timeval tv1, tv2; struct timezone tz; unsigned int *indices; while (argc-- > 0) { if ((*argv)[0] == '-') { /* look at first char of next */ switch ((*argv)[1]) { /* look at second */ case 'b': argv++; argc--; memsize = atoi(*argv) * 1024; break; case 'n': argv++; argc--; count = atoi(*argv); break; case 's': argv++; argc--; stride = atoi(*argv); break; default: usage(); exit(1); break; } } argv++; } char* mem = mmap(NULL, memsize, PROT_READ | PROT_WRITE, MAP_PRIVATE | MAP_ANON, -1, 0); // trick3: init pointer chasing, per stride=8 byte size = memsize / stride; indices = malloc(size * sizeof(int)); for (i = 0; i < size; i++) indices[i] = i; // trick 2: fill mem with pointer references for (i = 0; i < size - 1; i++) *(char **)&mem[indices[i]*stride]= (char*)&mem[indices[i+1]*stride]; *(char **)&mem[indices[size-1]*stride]= (char*)&mem[indices[0]*stride]; char **p = (char **) mem; tmp = count / 100; gettimeofday (&tv1, &tz); for (i = 0; i < tmp; ++i) { HUNDRED; //trick 1 } gettimeofday (&tv2, &tz); if (tv2.tv_usec < tv1.tv_usec) { usec = 1000000 + tv2.tv_usec - tv1.tv_usec; sec = tv2.tv_sec - tv1.tv_sec - 1; } else { usec = tv2.tv_usec - tv1.tv_usec; sec = tv2.tv_sec - tv1.tv_sec; } printf("Buffer size: %ld KB, stride %d, time %d.%06d s, latency %.2f ns\n", memsize/1024, stride, sec, usec, (sec * 1000000 + usec) * 1000.0 / (tmp *100)); munmap(mem, memsize); free(indices);}这里用到了 3 个小技巧：
...