C/C++性能提升之cache分析

在开发过程中，我们有时会碰到程序性能瓶颈，这时候需要我们查找热点代码，借用一些命令、工具去分析自己的程序，下面我就介绍一下如何使用perf工具分析程序的cache命中率。

在编写代码前先介绍一下我们的硬件平台，我电脑的CPU 是酷睿i7-12700h (14个核20线程),系统ubuntu22.04, 内存16G，大概信息如下：

接下来上我们的代码：

//
// Created by rookie on 23-6-4.
//
#define _GNU_SOURCE             /* See feature_test_macros(7) */
#include <sched.h>
#include <pthread.h>
#include <stdlib.h>
#include <stdio.h>
#include <sys/time.h>
#include <unistd.h>
#include <sys/syscall.h>   /* For SYS_xxx definitions */#if  0
struct multi_var
{volatile int threadAVar;volatile int threadBVar;
}g_multi_var __attribute__((aligned(64)));
#else
struct multi_var
{volatile int threadAVar __attribute__((aligned(64)));volatile int threadBVar __attribute__((aligned(64)));
}g_multi_var;
#endifstatic const char *usage_str = "test cache\r\n";
static void usage()
{printf("%s", usage_str);
}void test_cpu_performance(unsigned  int cpu)
{float f1 = 1.2;double db1 = 2.3;int i;int count = 0;int n = 0, n1 = 0;double d = 1;struct timeval a1, a2;gettimeofday(&a1, NULL);if(cpu == 0){count = 0;while (count++ <= 100){for (n = 1; n < 1024*1024; ++n) {g_multi_var.threadAVar += n;}}} else if(cpu == 1){count = 0;while (count++ <= 100){for (n = 1; n < 1024*1024; ++n) {g_multi_var.threadBVar += n;}}}gettimeofday(&a2, NULL);long time = 1000000*(a2.tv_sec-a1.tv_sec)+a2.tv_usec-a1.tv_usec;printf("cpu = %d,time = %ld exit\r\n", cpu, time);
}static int set_cpu_affinity(unsigned int cpu)
{cpu_set_t cpuset;pid_t tid = syscall(SYS_gettid);CPU_ZERO(&cpuset);CPU_SET(cpu, &cpuset);printf("before sched setaffinity cpu=%d\r\n", cpu);if(sched_setaffinity(tid, sizeof(cpuset), &cpuset) < 0){perror("sched setaffinity");printf("cpu %d\r\n", cpu);return -1;}printf("pid = %d, after sched setaffinity cpu=%d\r\n", tid, cpu);return 0;
}
static void* wast_cpu_body(void* arg)
{int i = 2;unsigned  int cpu = *(unsigned int *)arg;printf("wast_cpu_body in cpu=%d\r\n", cpu);set_cpu_affinity(cpu);test_cpu_performance(cpu);return NULL;
}int main(int argc, char* argv[])
{pthread_t *thread;int cpunum = 4;int ret = 0;int j = 0;int *icpu = NULL;pthread_attr_t *attr = NULL;void *p_arg = NULL;if(argc > 2){usage();}else if(argc == 2){cpunum = atoi(argv[1]);thread = malloc(sizeof(*thread) * cpunum);attr = malloc(sizeof(*attr) * cpunum);icpu = malloc(sizeof(*icpu) * cpunum);for (j = 0; j < cpunum; ++j) {ret = pthread_attr_init(&attr[j]);
//        ret |= pthread_attr_setdetachstate(&attr[j], PTHREAD_CREATE_DETACHED);icpu[j] = j;ret |= pthread_create(&thread[j], &attr[j], (void*(*)(void*))wast_cpu_body, &icpu[j]);printf("cpu index = %d\r\n", j);}for (j = 0; j < cpunum; ++j) {pthread_join(thread[j], NULL);}for (j = 0; j < cpunum; ++j) {pthread_attr_destroy(&attr[j]);}free(icpu);free(thread);free(attr);}return ret;
}

这里说明一下，关于代码中的__attribute__((aligned(64))) 对齐字节数为什么选择64？

这是由于我电脑的CPU的CACHE_LINE就是64，具体可以使用getconf命令查看

接下来我们编译出来两个可执行文件，test_cache0与test_cache1，与上述条件判断的对应关系如下：

//test_cache1 !!!!
struct multi_var
{volatile int threadAVar;volatile int threadBVar;
}g_multi_var __attribute__((aligned(64)));//test_cache0 !!!!
struct multi_var
{volatile int threadAVar __attribute__((aligned(64)));volatile int threadBVar __attribute__((aligned(64)));
}g_multi_var;

然后我们使用perf检测它们执行的情况

sudo perf stat -e L1-dcache-load-misses,L1-icache-load-misses,L1-dcache-load,L1-dcache-stores,branch-load-miss,branch-loads  ./test_cache0 10sudo perf stat -e L1-dcache-load-misses,L1-icache-load-misses,L1-dcache-load,L1-dcache-stores,branch-load-miss,branch-loads  ./test_cache1 10