我试图使用时间()来测量我的程序的各个点。

我不明白的是为什么前后的值是一样的?我知道这不是分析我的程序的最佳方式,我只是想看看需要多长时间。

printf("**MyProgram::before time= %ld\n", time(NULL));

doSomthing();
doSomthingLong();

printf("**MyProgram::after time= %ld\n", time(NULL));

我试过:

struct timeval diff, startTV, endTV;

gettimeofday(&startTV, NULL); 

doSomething();
doSomethingLong();

gettimeofday(&endTV, NULL); 

timersub(&endTV, &startTV, &diff);

printf("**time taken = %ld %ld\n", diff.tv_sec, diff.tv_usec);

我如何读取**时间花费= 0 26339的结果?这是否意味着26339纳秒= 26.3毫秒?

**时间= 4 45025,这是否意味着4秒25毫秒?


当前回答

下面是一个简单的类,它将在指定的持续时间单位内打印它进入和离开作用域之间的持续时间:

#include <chrono>
#include <iostream> 

template <typename T>
class Benchmark
{
   public:
    Benchmark(std::string name) : start(std::chrono::steady_clock::now()), name(name) {}
    ~Benchmark()
    {
        auto end = std::chrono::steady_clock::now();
        T duration = std::chrono::duration_cast<T>(end - start);
        std::cout << "Bench \"" << name << "\" took: " << duration.count() << " units" << std::endl;
    }

   private:
    std::string name;
    std::chrono::time_point<std::chrono::steady_clock> start;

};
int main()
{
  Benchmark<std::chrono::nanoseconds> bench("for loop");
  for(int i = 0; i < 1001000; i++){}
}

使用示例:

int main()
{
  Benchmark<std::chrono::nanoseconds> bench("for loop");
  for(int i = 0; i < 100000; i++){}
}

输出:

Bench "for loop" took: 230656 units

其他回答

仅限Windows:(Linux标签是在我发布这个答案后添加的)

您可以使用GetTickCount()来获取自系统启动以来所经过的毫秒数。

long int before = GetTickCount();

// Perform time-consuming operation

long int after = GetTickCount();
#include <ctime>

void f() {
  using namespace std;
  clock_t begin = clock();

  code_to_time();

  clock_t end = clock();
  double elapsed_secs = double(end - begin) / CLOCKS_PER_SEC;
}

time()函数只精确到一秒内,但是一秒内有CLOCKS_PER_SEC“时钟”。这是一种简单、便携的测量方法,尽管它过于简化了。

我通常使用以下方法:

#include <chrono>
#include <type_traits>

using perf_clock = std::conditional<
    std::chrono::high_resolution_clock::is_steady,
    std::chrono::high_resolution_clock,
    std::chrono::steady_clock
>::type;

using floating_seconds = std::chrono::duration<double>;

template<class F, class... Args>
floating_seconds run_test(Func&& func, Args&&... args)
{
   const auto t0 = perf_clock::now();
   std::forward<Func>(func)(std::forward<Args>(args)...);
   return floating_seconds(perf_clock::now() - t0);
} 

这与@nikos-athanasiou建议的相同,除了我避免使用非稳定时钟,并使用浮动秒数作为持续时间。

#include<time.h> // for clock
#include<math.h> // for fmod
#include<cstdlib> //for system
#include <stdio.h> //for delay

using namespace std;

int main()
{


   clock_t t1,t2;

   t1=clock(); // first time capture

   // Now your time spanning loop or code goes here
   // i am first trying to display time elapsed every time loop runs

   int ddays=0; // d prefix is just to say that this variable will be used for display
   int dhh=0;
   int dmm=0;
   int dss=0;

   int loopcount = 1000 ; // just for demo your loop will be different of course

   for(float count=1;count<loopcount;count++)
   {

     t2=clock(); // we get the time now

     float difference= (((float)t2)-((float)t1)); // gives the time elapsed since t1 in milliseconds

    // now get the time elapsed in seconds

    float seconds = difference/1000; // float value of seconds
    if (seconds<(60*60*24)) // a day is not over
    {
        dss = fmod(seconds,60); // the remainder is seconds to be displayed
        float minutes= seconds/60;  // the total minutes in float
        dmm= fmod(minutes,60);  // the remainder are minutes to be displayed
        float hours= minutes/60; // the total hours in float
        dhh= hours;  // the hours to be displayed
        ddays=0;
    }
    else // we have reached the counting of days
    {
        float days = seconds/(24*60*60);
        ddays = (int)(days);
        float minutes= seconds/60;  // the total minutes in float
        dmm= fmod(minutes,60);  // the rmainder are minutes to be displayed
        float hours= minutes/60; // the total hours in float
        dhh= fmod (hours,24);  // the hours to be displayed

    }

    cout<<"Count Is : "<<count<<"Time Elapsed : "<<ddays<<" Days "<<dhh<<" hrs "<<dmm<<" mins "<<dss<<" secs";


    // the actual working code here,I have just put a delay function
    delay(1000);
    system("cls");

 } // end for loop

}// end of main 
//***C++11 Style:***
#include <chrono>

std::chrono::steady_clock::time_point begin = std::chrono::steady_clock::now();
std::chrono::steady_clock::time_point end = std::chrono::steady_clock::now();

std::cout << "Time difference = " << std::chrono::duration_cast<std::chrono::microseconds>(end - begin).count() << "[µs]" << std::endl;
std::cout << "Time difference = " << std::chrono::duration_cast<std::chrono::nanoseconds> (end - begin).count() << "[ns]" << std::endl;