循环开销(loop overhead)测试程序

来源：互联网发布：nba2015常规赛mvp数据编辑：程序博客网时间：2024/05/18 01:52

同样的运算次数，for 循环的个数不同，开销也是不同的。

我表示很无知……求解释~

linux下代码：

#include <iostream>#include <sys/time.h>#include <unistd.h>using namespace std;int main() { long s; struct timeval start, end; for (int k = 0; k < 20; k++){//8gettimeofday(&start, NULL);s = 0;for (int i = 0; i < 10; i ++) {for (int j = 0; j < 10; j ++) {for (int l = 0; l < 10; l ++) {for (int m = 0; m < 10; m ++) {for (int n = 0; n < 10; n ++) {for (int o = 0; o < 10; o ++) {for (int p = 0; p < 10; p ++) {for (int q = 0; q < 10; q ++) {s++;}}}}}}}}gettimeofday(&end, NULL);double span = end.tv_sec - start.tv_sec + (end.tv_usec - start.tv_usec)/1000000.0;cout << span << endl;//4gettimeofday(&start, NULL);s = 0;for (int i = 0; i < 100; i ++) {for (int j = 0; j < 100; j ++) {for (int l = 0; l < 100; l ++) {for (int m = 0; m < 100; m ++) {s++;}}}}gettimeofday(&end, NULL);span = end.tv_sec - start.tv_sec + (end.tv_usec - start.tv_usec)/1000000.0;cout << span << endl; //2 gettimeofday(&start, NULL); s = 0; for (int i = 0; i < 10000; i ++) { for (int j = 0; j < 10000; j ++) { s++; } } gettimeofday(&end, NULL); span = end.tv_sec - start.tv_sec + (end.tv_usec - start.tv_usec)/1000000.0; cout << span << endl; //1 gettimeofday(&start, NULL); s = 0; for (int i = 0; i < 100000000; i ++) { s++; } gettimeofday(&end, NULL); span = end.tv_sec - start.tv_sec + (end.tv_usec - start.tv_usec)/1000000.0; cout << span << endl; }return 0;}

结果：

111

分析：同样的运算量，for循环越多，耗时越长。循环体为空，效果更明显