Question

循环访问QueryPerformanceCounter（）并保存值：

// Main loop for timer test
for ( int i = 0; i < ITERATIONS; i++ ) // ITERATIONS = 1000
{
    QueryPerformanceCounter(&li);
    time[i] = double(li.QuadPart) / PCFreq; //1,193,182 per second
}
//calculate the difference between each call 
// and save in difference[]
for ( int j = 0; j < (ITERATIONS - 1)  ; j++ )
{
    difference[j] = time[j+1] - time[j];
}

（除以PCFreq给出每次通话之间的时间。）

高分辨率计时器/计数器应该正常工作，因为它没有返回默认频率1000.

每个时间戳之间的平均值为11.990884微秒（一千个时间戳记）。

这似乎非常缓慢。

这个测试有缺陷吗？

或者为什么在1.1Ghz Celeron上报告这么慢的值？

Answer 1

在第一个循环中消除浮点数学可能是值得的，以便不考虑Win 7 Desktop和Embedded Compact 7之间的（潜在）差异。所以，像：

LARGE_INTEGER counter[ITERATIONS];
// Main loop for timer test
for ( int i = 0; i < ITERATIONS; i++ ) // ITERATIONS = 1000
{
    QueryPerformanceCounter(&counter[i]);
}
time[0] = double(counter[0].QuadPart) / PCFreq; //1,193,182 per second
//calculate the difference between each call 
// and save in difference[]
for ( int j = 0; j < (ITERATIONS - 1)  ; j++ )
{
    time[j+1] = double(counter[j+1].QuadPart) / PCFreq; //1,193,182 per second
    difference[j] = time[j+1] - time[j];
}

QueryPerformanceCounter（）测试Windows Embedded Compact 7

1 个答案: