time(NULL)函数调用将返回自epoc: 1970年1月1日以来经过的秒数。也许你要做的是取两个时间戳的差值:

size_t start = time(NULL);
doSomthing();
doSomthingLong();

printf ("**MyProgram::time elapsed= %lds\n", time(NULL) - start);

#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 

从所看到的内容来看，tv_sec存储所经过的秒数，而tv_usec单独存储所经过的微秒数。它们不是彼此的转换。因此，必须将它们更改为适当的单位，并添加它们以获得所消耗的总时间。

struct timeval startTV, endTV;

gettimeofday(&startTV, NULL); 

doSomething();
doSomethingLong();

gettimeofday(&endTV, NULL); 

printf("**time taken in microseconds = %ld\n",
    (endTV.tv_sec * 1e6 + endTV.tv_usec - (startTV.tv_sec * 1e6 + startTV.tv_usec))
    );

我已经创建了一个类来自动测量流逝的时间，请检查代码(c++11)在这个链接:https://github.com/sonnt174/Common/blob/master/time_measure.h

使用timmeasure类的示例:

void test_time_measure(std::vector<int> arr) {
  TimeMeasure<chrono::microseconds> time_mea;  // create time measure obj
  std::sort(begin(arr), end(arr));
}

Matlab味!

Tic启动一个秒表来测量性能。该功能记录执行tic命令时的内部时间。使用toc函数显示运行时间。

#include <iostream>
#include <ctime>
#include <thread>
using namespace std;

clock_t START_TIMER;

clock_t tic()
{
    return START_TIMER = clock();
}

void toc(clock_t start = START_TIMER)
{
    cout
        << "Elapsed time: "
        << (clock() - start) / (double)CLOCKS_PER_SEC << "s"
        << endl;
}

int main()
{
    tic();
    this_thread::sleep_for(2s);
    toc();

    return 0;
}

#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。