我似乎无法得到一个简单的程序（有大量的内存访问），以实现在Linux一致的时间。 我正在使用2.6内核，并且该程序正在具有实时优先级的双核处理器上运行。 我试图通过将内存数组声明为volatile来禁用caching效果。 以下是结果和程序。 exception值有哪些可能的来源？
结果：
Number of trials: 100 Range: 0.021732s to 0.085596s Average Time: 0.058094s Standard Deviation: 0.006944s Extreme Outliers (2 SDs away from mean): 7 Average Time, excluding extreme outliers: 0.059273s 程序：
#include <stdio.h> #include <stdlib.h> #include <math.h> #include <sched.h> #include <sys/time.h> #define NUM_POINTS 5000000 #define REPS 100 unsigned long long getTimestamp() { unsigned long long usecCount; struct timeval timeVal; gettimeofday(&timeVal, 0); usecCount = timeVal.tv_sec * (unsigned long long) 1000000; usecCount += timeVal.tv_usec; return (usecCount); } double convertTimestampToSecs(unsigned long long timestamp) { return (timestamp / (double) 1000000); } int main(int argc, char* argv[]) { unsigned long long start, stop; double times[REPS]; double sum = 0; double scale, avg, newavg, median; double stddev = 0; double maxval = -1.0, minval = 1000000.0; int i, j, freq, count; int outliers = 0; struct sched_param sparam; sched_getparam(getpid(), &sparam); sparam.sched_priority = sched_get_priority_max(SCHED_FIFO); sched_setscheduler(getpid(), SCHED_FIFO, &sparam); volatile float* data; volatile float* results; data = calloc(NUM_POINTS, sizeof(float)); results = calloc(NUM_POINTS, sizeof(float)); for (i = 0; i < REPS; ++i) { start = getTimestamp(); for (j = 0; j < NUM_POINTS; ++j) { results[j] = data[j]; } stop = getTimestamp(); times[i] = convertTimestampToSecs(stop-start); } free(data); free(results); for (i = 0; i < REPS; i++) { sum += times[i]; if (times[i] > maxval) maxval = times[i]; if (times[i] < minval) minval = times[i]; } avg = sum/REPS; for (i = 0; i < REPS; i++) stddev += (times[i] - avg)*(times[i] - avg); stddev /= REPS; stddev = sqrt(stddev); for (i = 0; i < REPS; i++) { if (times[i] > avg + 2*stddev || times[i] < avg - 2*stddev) { sum -= times[i]; outliers++; } } newavg = sum/(REPS-outliers); printf("Number of trials: %d\n", REPS); printf("Range: %fs to %fs\n", minval, maxval); printf("Average Time: %fs\n", avg); printf("Standard Deviation: %fs\n", stddev); printf("Extreme Outliers (2 SDs away from mean): %d\n", outliers); printf("Average Time, excluding extreme outliers: %fs\n", newavg); return 0; }
确保没有其他进程需要CPU时间。 特别注意屏幕保护程序和任何定期更新GUI（例如时钟或类似）的东西。 尝试为基准进程设置CPU关联，将其锁定到一个核心（例如命令行中的任务集）。 如果不是分页，则进行基准测试 – 通常，您希望有一个外部循环运行N次，然后执行最后一次N-1执行。