windows下的精確計時函數

BOOL QueryPerformanceFrequency(LARGE_INTEGER *lpFrequency)獲取系統的計數器的頻率

BOOL QueryPerformanceCounter(LARGE_INTEGER *lpPerformanceCount)獲取計數器的值

(原文出處：coder_xia的博客：http://blog.csdn.net/coder_xia/article/details/6566708)

   然後用兩次計數器的差除以Frequency就得到時間。

測試程序如下：

#include <iostream>
#include <windows.h>
using namespace std;
int main()
{
    LARGE_INTEGER m_nFreq;
    LARGE_INTEGER m_nBeginTime;
    LARGE_INTEGER nEndTime;
    QueryPerformanceFrequency(&m_nFreq); // 獲取時鐘週期
    QueryPerformanceCounter(&m_nBeginTime); // 獲取時鐘計數
    Sleep(100);
    QueryPerformanceCounter(&nEndTime);
    cout << (double)(nEndTime.QuadPart-m_nBeginTime.QuadPart)*1000/m_nFreq.QuadPart << endl;
}

 

需要注意的就是結果需要強制轉換爲double，不然會得到如下錯誤：<< is ambiguous