通过perf_event函数获取ipc

最新推荐文章于 2025-05-14 14:15:27 发布

原创最新推荐文章于 2025-05-14 14:15:27 发布 · 2k 阅读

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本文提供了一个使用perf_event_open函数测量特定进程的CPU周期和指令数的C语言示例程序。该程序首先打开两个硬件性能计数器，分别用于统计CPU周期和指令数，并通过两次读取来计算间隔内的变化，最终输出每周期指令数(IPC)。

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其实这个问题困扰了蛮久，主要还是不太会用..最后参考git上一个代码搞定了，还是很开心的。https://gist.github.com/teqdruid/2473071

直接上代码吧：

#include <stdlib.h>
#include <stdio.h>
#include <unistd.h>
#include <string.h>
#include <sys/ioctl.h>
#include <linux/perf_event.h>
#include <asm/unistd.h>

#include <time.h>
#include <fcntl.h>
#include <sys/file.h>
#include <sys/time.h>
#include <assert.h>
#include <stdint.h>
#include <sys/syscall.h>
#include <linux/unistd.h>



#define PID_NUM 176139
int Core_id=-1;

static long
perf_event_open(struct perf_event_attr *hw_event, pid_t pid,
                int cpu, int group_fd, unsigned long flags)
{
    int ret;

    ret = syscall(__NR_perf_event_open, hw_event, pid, cpu,
                   group_fd, flags);
    return ret;
}


// Setup info for perf_event
struct perf_event_attr attr[2];


int
main(int argc, char **argv)
{
int pid,core_id,i,rc;
    uint64_t val1[2],val2[2];
    if(argc==1) {
        printf("please add pid number\n");
        return -1;
    }
    for(pid=i=0;argv[1][i]!=0;i++){
        if(argv[1][i]<'0'||argv[1][i]>'9') {
        printf("illegal pid number\n");
        return -1;
        }
        pid=pid*10+argv[1][i]-'0';
    }
    printf("%d\n",pid);
    struct perf_event_attr pe;
    long long count,cycles,instructions;
    double ipc;
    int fd[2];

    core_id=Core_id;
    cycles=instructions=0;
    memset(&pe, 0, sizeof(struct perf_event_attr));

    attr[0].type = PERF_TYPE_HARDWARE;
    attr[0].config = PERF_COUNT_HW_CPU_CYCLES; /* generic PMU event*/
    attr[0].disabled = 0;
    fd[0] = perf_event_open(&attr[0], pid , core_id, -1, 0);
    if (fd[0] < 0) {
            perror("Opening performance counter");
    }

    attr[1].type = PERF_TYPE_HARDWARE;
    attr[1].config = PERF_COUNT_HW_INSTRUCTIONS; /* generic PMU event*/
    attr[1].disabled = 0;
    fd[1] = perf_event_open(&attr[1], pid , core_id, -1, 0);
    if (fd[1] < 0) {
            perror("Opening performance counter");
    }


    //count cycles,instructions;
    rc = read(fd[0], &val1[0], sizeof(val1[0]));  assert(rc);
    rc = read(fd[1], &val1[1], sizeof(val1[1]));  assert(rc);
    cycles= val1[0];
    instructions = val1[1];
    printf("round 1:cycles =  %lld instructions=  %lld\n",val1[0],val1[1]);

    usleep(100000);//usleep takes microsecond

    rc = read(fd[0], &val1[0], sizeof(val1[0]));  assert(rc);
    rc = read(fd[1], &val1[1], sizeof(val1[1]));  assert(rc);
    printf("round 2:cycles =  %lld instructions=  %lld\n",val1[0],val1[1]);
    cycles= val1[0]-cycles;
    instructions = val1[1]-instructions;
    ipc=(double)instructions/(double)cycles;
    printf("cycles =  %lld  instructions =  %lld  IPC=%f\n ",cycles ,instructions,ipc);

    close(fd[0]);
    close(fd[1]);
    return 0 ;

}

gcc test.c - o xxx

./xxx pid

其中有个坑，我们都知道在命令行下sleep可以用小数表示，比如sleep(0.5)表示0.5秒。

但是这里在C语言中，是整数的，用了小数虽然可以编译通过，都取整为0，那就直接连续两次读取寄存器信息，没有延迟效果，结果自然就很不准确。

换成了usleep()函数即可解决，usleep单位为微妙。

perf_event_open()函数介绍可以参考http://man7.org/linux/man-pages/man2/perf_event_open.2.html