C语言程序函数计时仅显示三位小数的原因排查求助

问题描述 #

我在编写测试不同排序算法的C程序时，发现计时输出只能显示三位小数（如CSV中0.001000、0.002000这类格式），即使使用%f格式输出也是如此。使用环境为Windows 11系统、AMD Ryzen 5处理器，曾怀疑是硬件或系统配置导致。

相关代码片段 #

code.c #

// insertion sort test
time_t t = 0;
t = clock();
insertion(arr, n);
t = clock() - t;
double time_taken = ((double)t)/CLOCKS_PER_SEC; // in seconds
printf("insertion() took %f seconds to execute\n", time_taken);
fprintf(fp, "%f,", time_taken);
fflush(fp);

data.csv #

0.001000,0.000000,0.000000,0.001000,0.000000
0.002000,0.000000,0.000000,0.002000,0.000000
0.005000,0.000000,0.000000,0.005000,0.000000
0.008000,0.001000,0.000000,0.008000,0.000000
0.004000,0.000000,0.000000,0.021000,0.000000
0.011000,0.000000,0.000000,0.024000,0.001000
0.008000,0.016000,0.000000,0.025000,0.000000
0.031000,0.001000,0.001000,0.029000,0.000000
0.041000,0.001000,0.000000,0.033000,0.000000
0.060000,0.001000,0.001000,0.051000,0.001000
0.062000,0.001000,0.000000,0.059000,0.001000
0.074000,0.002000,0.001000,0.068000,0.000000
0.085000,0.002000,0.001000,0.080000,0.000000

原因分析 #

这不是硬件或系统配置的问题，而是clock()函数的精度限制：

• Windows平台下，CLOCKS_PER_SEC被定义为1000，意味着clock()的计时精度是毫秒级（1/1000秒）。
• 计算出的time_taken最小单位就是0.001秒，所以输出只会以0.001的整数倍显示，看起来只有三位有效小数。

解决方案 #

1. 使用高精度Windows API计时 #

改用QueryPerformanceCounter和QueryPerformanceFrequency，这两个函数可以提供微秒甚至纳秒级的精度，完全满足细粒度计时需求：

#include <windows.h>

// 高精度计时示例
LARGE_INTEGER freq, start, end;
QueryPerformanceFrequency(&freq); // 获取计数器频率
QueryPerformanceCounter(&start);  // 记录开始时间

insertion(arr, n); // 执行排序函数

QueryPerformanceCounter(&end);    // 记录结束时间
double time_taken = (double)(end.QuadPart - start.QuadPart) / freq.QuadPart;

printf("insertion() took %.6f seconds to execute\n", time_taken);
fprintf(fp, "%.6f,", time_taken);
fflush(fp);

2. 多次运行取平均值（兼容原代码） #

如果不想修改计时逻辑，可以多次运行排序函数，把总时间除以运行次数，放大时间基数以获得更精确的单次时间：

int runs = 1000;
time_t t = clock();
for (int i = 0; i < runs; i++) {
    insertion(arr, n);
}
t = clock() - t;
double time_taken = ((double)t)/CLOCKS_PER_SEC / runs; // 单次运行时间

printf("insertion() took %.6f seconds to execute\n", time_taken);
fprintf(fp, "%.6f,", time_taken);
fflush(fp);

内容的提问来源于stack exchange，提问作者Grovetender