Easily measure elapsed time

Question

I am trying to use time() to measure various points of my program.

What I don\'t understand is why the values in the before and after are the same? I understand thi

Answer 1

On linux, clock_gettime() is one of the good choices. You must link real time library(-lrt).

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <time.h>

#define BILLION  1000000000L;

int main( int argc, char **argv )
  {
    struct timespec start, stop;
    double accum;

    if( clock_gettime( CLOCK_REALTIME, &start) == -1 ) {
      perror( "clock gettime" );
      exit( EXIT_FAILURE );
    }

    system( argv[1] );

    if( clock_gettime( CLOCK_REALTIME, &stop) == -1 ) {
      perror( "clock gettime" );
      exit( EXIT_FAILURE );
    }

    accum = ( stop.tv_sec - start.tv_sec )
          + ( stop.tv_nsec - start.tv_nsec )
            / BILLION;
    printf( "%lf\n", accum );
    return( EXIT_SUCCESS );
  }

Answer 2

//***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;

Answer 3

time(NULL) returns the number of seconds elapsed since 01/01/1970 at 00:00 (the Epoch). So the difference between the two values is the number of seconds your processing took.

int t0 = time(NULL);
doSomthing();
doSomthingLong();
int t1 = time(NULL);

printf ("time = %d secs\n", t1 - t0);

You can get finer results with getttimeofday(), which return the current time in seconds, as time() does and also in microseconds.

Answer 4

As I can see from your question, it looks like you want to know the elapsed time after execution of some piece of code. I guess you would be comfortable to see the results in second(s). If so, try using difftime() function as shown below. Hope this solves your problem.

#include <time.h>
#include <stdio.h>

time_t start,end;
time (&start);
.
.
.
<your code>
.
.
.
time (&end);
double dif = difftime (end,start);
printf ("Elasped time is %.2lf seconds.", dif );

Answer 5

Internally the function will access the system's clock, which is why it returns different values each time you call it. In general with non-functional languages there can be many side effects and hidden state in functions which you can't see just by looking at the function's name and arguments.

Answer 6

#include <ctime>
#include <functional>

using namespace std;

void f() {
  clock_t begin = clock();

  // ...code to measure time...

  clock_t end = clock();

  function<double(double, double)> convtime = [](clock_t begin, clock_t end)
  {
     return double(end - begin) / CLOCKS_PER_SEC;
  };

  printf("Elapsed time: %.2g sec\n", convtime(begin, end));

}

Similar example to one available here, only with additional conversion function + print out.