|
LARGE_INTEGER litmp;
LONGLONG QPart1,QPart2;;
double dfMinus, dfFreq, dfTim;
QueryPerformanceFrequency(&litmp);
dfFreq = (double)litmp.QuadPart; // 获得计数器的时钟频率
QueryPerformanceCounter(&litmp);
QPart1 = litmp.QuadPart; // 获得初始值
QPart2=QPart1;
while(TRUE)
{
QPart1 = QPart2;// 上一次的终止值变成新的起始值
do{
QueryPerformanceCounter(&litmp);
QPart2 = litmp.QuadPart;// 获得中止值
dfMinus = (double)(QPart2-QPart1);
dfTim = dfMinus / dfFreq; // 获得对应的时间值,单位为秒
}while(dfTim<0.001);//
}
请教这段代码在2440上能否起到1ms的定时 |
|