Question

这是我尝试获取资源的简单方法，特别是在程序执行时发生的总上下文切换。

#include<stdio.h>
#include <sys/resource.h>

int appgetrusage(struct rusage *);
int appgetdiffrusage(struct rusage *, struct rusage *);

int main() {
    struct rusage begin, end;

    appgetrusage(&begin);

/*
* core of the program goes here 
* where lot of system threads are spawned and joined
*
*/

    appgetrusage(&end);
    appgetdiffrusage(&begin, &end);

    return 0;
}


int appgetrusage(struct rusage *usage){
    int who = RUSAGE_SELF;
    struct timeval start, end;
    getrusage(RUSAGE_SELF, usage);
    return 1;
}


int appgetdiffrusage(struct rusage *oldr, struct rusage *newr){

    printf("\n");    
    printf("user time (ms): %llu\n",1000 * ((newr->ru_utime).tv_sec - (oldr->ru_utime).tv_sec) + 
            ((newr->ru_utime).tv_usec - (oldr->ru_utime).tv_usec) / 1000);

    printf("system time (ms): %ld\n", 1000 * ((newr->ru_stime).tv_sec - (oldr->ru_stime).tv_sec) + 
            ((newr->ru_stime).tv_usec - (oldr->ru_stime).tv_usec) / 1000);         

    printf("voluntary context switches : %ld\n", newr->ru_nvcsw - oldr->ru_nvcsw);
    printf("involuntary context switches : %ld\n", newr->ru_nivcsw - oldr->ru_nivcsw);

    return 1;
}

这是正确的方法吗？
有人可以建议替代方案吗？或更正？： - ）

Answer 1

有人可以建议替代方案吗？

使用perf（https://perf.wiki.kernel.org/index.php/Tutorial#Counting_with_perf_stat）：

perf stat your-program

像这样：

>perf stat ./my_test 2
Thread 139828421826304:
Thread 139828411336448:
^[[A./my_test: Terminated

 Performance counter stats for './my_test 2':

      74333.536760 task-clock                #    1.999 CPUs utilized
               627 context-switches          #    0.008 K/sec
                26 cpu-migrations            #    0.000 K/sec
               282 page-faults               #    0.004 K/sec
      182727508914 cycles                    #    2.458 GHz                     [50.00%]
   <not supported> stalled-cycles-frontend
   <not supported> stalled-cycles-backend
      121168605770 instructions              #    0.66  insns per cycle         [75.00%]
       30262379463 branches                  #  407.116 M/sec                   [74.99%]
           1635031 branch-misses             #    0.01% of all branches         [75.01%]

      37.181359478 seconds time elapsed

Answer 2

Systemtap为分析/调试用户空间以及内核代码提供了出色的工具。

https://access.redhat.com/site/documentation/en-US/Red_Hat_Enterprise_Linux/5/html/SystemTap_Beginners_Guide/

systemtap脚本，用于查找上下文切换 https://sourceware.org/systemtap/examples/process/chng_cpu.stp

在ARM中，由于uprobes（https://blueprints.launchpad.net/linux-linaro/+spec/arm-uprobes），用户空间分析尚不可用。但是X86，它运行正常。

Answer 3

你们已经分享了最好的资源，比如perf和systemtap，但是如果有人只是不想深入了解那么深入了解RHEL6或内核2.6.23及更高版本的sysstat包就会有一些叫做pidstat的东西。将pidstat与-w一起使用，它将为您提供自愿和非自愿上下文切换的总数。

在C / C ++应用程序中测量总上下文切换的最佳方法是什么？

3 个答案: