本文介绍了C `clock()` 函数只返回一个零的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着跟版网的小编来一起学习吧！
问题描述
C clock() 函数只返回零.我尝试使用不同的类型，但没有任何改进......这是一种以高精度测量时间的好方法吗?
#include #include int main(){clock_t 开始，结束；双 cpu_time_used;字符 [32];开始 = 时钟()；printf("
睡眠 3 秒...
");睡眠(3)；结束 = 时钟()；cpu_time_used = ((double)(end - start))/((double)CLOCKS_PER_SEC);printf("start = %.20f
end = %.20f
", start, end);printf("delta = %.20f
", ((double) (end - start)));printf("cpu_time_used = %.15f
", cpu_time_used);printf("CLOCKS_PER_SEC = %i
", CLOCKS_PER_SEC);返回0；} <块引用>
睡眠3秒...开始 = 0.00000000000000000000结束 = 0.00000000000000000000增量 = 0.00000000000000000000cpu_time_used = 0.000000000000000CLOCKS_PER_SEC = 1000000平台:Intel 32 位，RedHat Linux，gcc 3.4.6
解决方案
clock() 报告 CPU 使用时间.sleep() 不使用任何 CPU 时间.所以你的结果可能完全正确，只是不是你想要的.
The C clock() function just returns me a zero. I tried using different types, with no improvement... Is this a good way to measure time with good precision?
#include <time.h>
#include <stdio.h>
int main()
{
clock_t start, end;
double cpu_time_used;
char s[32];
start = clock();
printf("
Sleeping 3 seconds...
");
sleep(3);
end = clock();
cpu_time_used = ((double)(end - start)) / ((double)CLOCKS_PER_SEC);
printf("start = %.20f
end = %.20f
", start, end);
printf("delta = %.20f
", ((double) (end - start)));
printf("cpu_time_used = %.15f
", cpu_time_used);
printf("CLOCKS_PER_SEC = %i
", CLOCKS_PER_SEC);
return 0;
}
Sleeping 3 seconds... start = 0.00000000000000000000 end = 0.00000000000000000000 delta = 0.00000000000000000000 cpu_time_used = 0.000000000000000 CLOCKS_PER_SEC = 1000000
Platform: Intel 32 bit, RedHat Linux, gcc 3.4.6
解决方案
clock() reports CPU time used. sleep() doesn't use any CPU time. So your result is probably exactly correct, just not what you want.
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