测量时间结果返回值为0或0.001 [英] Measuring time results in return values of 0 or 0.001

问题描述

我试图使用 chrono :: steady_clock 来测量程序中代码块之间的小数秒。我有这个代码块在LiveWorkSpace工作（ http://liveworkspace.org/code/YT1I$9）：

  #include< chrono> 
 #include< iostream> 
 #include< vector> 
 
 int main（）
 {
 auto start = std :: chrono :: steady_clock :: now（）; 
 for（unsigned long long int i = 0; i <10000; ++ i）{
 std :: vector< int> v（i，1）; 
} 
 auto end = std :: chrono :: steady_clock :: now（）; 
 
 auto difference = std :: chrono :: duration_cast< std :: chrono :: microseconds>（end-start）.count（）; 
 
 std :: cout<< 自开始后的秒数：< （（双）差/ 1000000）;当我在我的程序中实现相同的想法，像这样：
 
 
 
 $ b < 
 
 
  auto start = std :: chrono :: steady_clock :: now（）; 
 //代码块到时间
 auto end = std :: chrono :: stead_clock :: now（）; 
 
 auto difference = std :: chrono :: duration_cast< std :: chrono :: microseconds>（end-start）.count（）
 
 std :: cout< < 自开始后的秒数：< （（双）差/ 1000000）; 
  
程序只打印出 0 和 0.001 。我非常怀疑我的代码块的执行时间总是等于 0 或 1000 微秒，这个舍入和如何消除它，以便我可以得到正确的小数值？
 
 
 
这是一个Windows程序。
解决方案
在MSVC2012上运行一些测试后，我可以确认Microsoft的实现中的C ++ 11时钟没有足够高的分辨率。有关此问题的错误报告，请参见 C ++标题的high_resolution_clock不具有高分辨率 。
 
 
 
因此，不幸的是，对于更高分辨率的计时器，您需要使用 boost :: chrono 或QueryPerformanceCounter像这样，直到他们修复错误：
  #include< iostream> 
 #include< Windows.h> 
 
 int main（）
 {
 LARGE_INTEGER frequency; 
 QueryPerformanceFrequency（& frequency）; 
 
 LARGE_INTEGER start; 
 QueryPerformanceCounter（& start）; 
 
 //将代码放入时间
 
 LARGE_INTEGER end; 
 QueryPerformanceCounter（& end）; 
 
 //微秒使用1000000.0 
 double interval = static_cast< double>（end.QuadPart- start.QuadPart）/ 
 frequency.QuadPart; // in seconds 
 std :: cout<<间隔; 
} 
  
 
I am trying to use chrono::steady_clock to measure fractional seconds elapsed between a block of code in my program. I have this block of code working in LiveWorkSpace (http://liveworkspace.org/code/YT1I$9):
#include <chrono>
#include <iostream>
#include <vector>

int main()
{
    auto start = std::chrono::steady_clock::now();
    for (unsigned long long int i = 0; i < 10000; ++i) {
       std::vector<int> v(i, 1);
    }
    auto end = std::chrono::steady_clock::now();

    auto difference = std::chrono::duration_cast<std::chrono::microseconds>(end - start).count();

    std::cout << "seconds since start: " << ((double)difference / 1000000);
}
When I implement the same idea into my program like so:
auto start = std::chrono::steady_clock::now();
// block of code to time
auto end = std::chrono::stead_clock::now();

auto difference = std::chrono::duration_cast<std::chrono::microseconds>(end - start).count()

std::cout << "seconds since start: " << ((double) difference / 1000000);
The program will only print out values of 0 and 0.001. I highly doubt that the execution time for my block of code always equals 0 or 1000 microseconds, so what is accounting for this rounding and how might I eliminate it so that I can get the proper fractional values?

This is a Windows program.
解决方案
After running some tests on MSVC2012, I could confirm that the C++11 clocks in Microsoft's implementation do not have a high enough resolution. See C++  header's high_resolution_clock does not have high resolution for a bug report concerning this issue.

So, unfortunately for a higher resolution timer, you will need to use boost::chrono or QueryPerformanceCounter directly like so until they fix the bug:
#include <iostream>
#include <Windows.h>

int main()
{
    LARGE_INTEGER frequency;
    QueryPerformanceFrequency(&frequency);

    LARGE_INTEGER start;
    QueryPerformanceCounter(&start);

    // Put code here to time

    LARGE_INTEGER end;
    QueryPerformanceCounter(&end);

    // for microseconds use 1000000.0
    double interval = static_cast<double>(end.QuadPart- start.QuadPart) / 
                      frequency.QuadPart; // in seconds
    std::cout << interval;
}   


                        
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