以微秒为单位计时函数
大家好，我正在尝试以微秒为单位计算我编写的一些搜索函数的时间，并且需要足够长的时间才能使其显示 2 位有效数字。我编写了这段代码来计时我的搜索功能，但它似乎运行得太快了。我总是得到 0 微秒，除非我运行搜索 5 次然后得到 1,000,000 微秒。我想知道我是否在数学上错误地得到了以微秒为单位的时间，或者是否有某种格式化函数可以用来强制它显示两个数字？
clock_t start = clock();
index = sequentialSearch.Sequential(TO_SEARCH);
index = sequentialSearch.Sequential(TO_SEARCH);
clock_t stop = clock();
cout << "number found at index " << index << endl;
int time = (stop - start)/CLOCKS_PER_SEC;
time = time * SEC_TO_MICRO;
cout << "time to search = " << time<< endl;
Hey guys I'm trying to time some search functions I wrote in microseconds, and it needs to take long enough to get it to show 2 significant digits. I wrote this code to time my search function but it seems to go too fast. I always end up getting 0 microseconds unless I run the search 5 times then I get 1,000,000 microseconds. I'm wondering if I did my math wrong to get the time in micro seconds, or if there's some kind of formatting function I can use to force it to display two sig figs?
clock_t start = clock();
index = sequentialSearch.Sequential(TO_SEARCH);
index = sequentialSearch.Sequential(TO_SEARCH);
clock_t stop = clock();
cout << "number found at index " << index << endl;
int time = (stop - start)/CLOCKS_PER_SEC;
time = time * SEC_TO_MICRO;
cout << "time to search = " << time<< endl;
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您在这一行中使用整数除法：
我建议使用
double或float类型，并且您可能需要转换除法的组件。You are using integer division on this line:
I suggest using a
doubleorfloattype, and you'll likely need to cast the components of the division.使用 QueryPerformanceCounter 和 QueryPerformanceFrequency，假设您在 Windows 平台上，
这里有一个指向 ms KB 如何使用 QueryPerformanceCounter 来计时代码
Use QueryPerformanceCounter and QueryPerformanceFrequency, assuming your on windows platform
here a link to ms KB How To Use QueryPerformanceCounter to Time Code