当 CPU 频率可变时，基于时钟的计时可靠吗？
测量经过时间的常见方法是：
const clock_t START = clock();
// ...
const clock_t END = clock();
double T_ELAPSED = (double)(END - START) / CLOCKS_PER_SEC;
我知道这不是测量实时时间的最佳方法，但我想知道它是否适用于具有可变频率 CPU 的系统。难道只是错误的吗？
A common way to measure elapsed time is:
const clock_t START = clock();
// ...
const clock_t END = clock();
double T_ELAPSED = (double)(END - START) / CLOCKS_PER_SEC;
I know this is not the best way to measure real time, but I wonder if it works on a system with a variable frequency CPU. Is it just wrong?
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有些系统架构可以改变 CPU 的频率，但具有单独且恒定的频率来驱动系统时钟。人们可能会认为 Clock() 函数会返回与 CPU 频率无关的时间，但这必须在代码要运行的每个系统上进行验证。
There are system architectures that change the frequency of the CPU but have a separate and constant frequency to drive a system clock. One would think that a
clock()function would return a time independent of the CPU frequency but this would have to be verified on each system the code is intended to run on.在可变主频 CPU 上使用并不好。
http://support.ntp.org/bin/view/Support/KnownHardwareIssues
Linux 上的 NTP（网络时间协议）守护进程存在问题。
大多数操作系统都有一些 API 调用来获取更准确的值，例如 Windows 上的 QueryPerformanceCounter
http://msdn.microsoft.com/en-us/library/ms644904%28VS.85%29.aspx
It is not good to use on a variable clock speed CPU.
http://support.ntp.org/bin/view/Support/KnownHardwareIssues
NTP (network time prototcol) daemon on linux had issues with it.
Most OS's have some API calls for more accurate values, an example being on windows, QueryPerformanceCounter
http://msdn.microsoft.com/en-us/library/ms644904%28VS.85%29.aspx