C ++获取更精确的tickcount值

Question

我使用Tick Count：

计算c ++中某些操作的性能

DWORD tstart = GetTickCount();                  
{                                           
    //...Some Operation...
}                                           
DWORD tend = GetTickCount();                        
double tt = (double) (tend-tstart)/(double) 1000;   
std::cout << "Exec Time: " << tt << " Seconds" << std::endl;

然而，这只给我值＆gt; = 0.015 ...

低于0.015的任何值似乎都打印为0 ..

还有其他方法可以获得更高的精确度吗？

Answer 1

使用std::chrono::high_resolution_clock获得最高分辨率

typedef std::chrono::high_resolution_clock Clock;
auto start = Clock::now();
... do some work ...
auto stop = Clock::now();

std::cout << stop-start << '\n';

Answer 2

使用cout格式化程序。

include <iomanip>

std::cout << "Exec Time: " << std::setprecision(12) << tt << " Seconds" << std::endl;

http://www.cplusplus.com/reference/iomanip/setprecision/

我建议在C ++ 11中使用 chrono new。

#include <chrono>
#include <iomanip>

std::chrono::system_clock::time_point begin_time = std::chrono::system_clock::now();

... work ...

std::chrono::system_clock::time_point end_time = std::chrono::system_clock::now();
long long elapsed_seconds = std::chrono::duration_cast<std::chrono::seconds>(end_time-begin_time).count();
std::cout << "Duration (min:seg): " << std::setw(2) << std::setfill('0') << (elapsed_seconds / 60) << ":" << std::setw(2) << std::setfill('0') << (elapsed_seconds % 60) << std::endl;