clock()

Hi I got some problem with the clock fct.

Code:

---

#include <time.h>
 #include <stdio.h>

int main(int argc, char *argv[])
{
      clock_t start, end;
      double cpu_time_used;
 
      start = clock();

              int i = 0;
        for(i= 0;i++;i<500);
       
      end = clock();
      cpu_time_used = ((double) (end - start)) / CLOCKS_PER_SEC;

        printf("start: %e\n", start);
        printf("end: %e\n",end);
       
}

}

---

Why do I get start and end to be equal?


And if I add printf("end: %e\n",cpu_time_used); in the end I get start, end and cpu_time_used all equal to zero.
I don't get it! *smash head on the wall*

Quote:

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In the GNU system, `clock_t' is equivalent to `long int'
Info: (libc.info.gz) Processor And CPU Time

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But when I write

Code:

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printf("start: %ld\n", start);

---

I get start = 0;
Why is that?

/CLK_TCK;

Or whatever your compiler supports. Basically, the number of ticks per second, on your system, for this timer.

A clock-tick is often 10ms, so anything shorter than that will register as "same".

Chances are also that the compiler completely removes the loop altogether, since it doesn't actually produce anything!

--
Mats

Quote:

---

Originally Posted by Adak

/CLK_TCK;

Or whatever your compiler supports. Basically, the number of ticks per second, on your system, for this timer.

---

You have to use CLOCKS_PER_SEC instead of CLK_TCK. According to
/usr/include/time.h, CLK_TCK is the "obsolete POSIX.1-1988 name"
for CLOCKS_PER_SEC.

Quote:

---

Originally Posted by matsp

A clock-tick is often 10ms, so anything shorter than that will register as "same".

Chances are also that the compiler completely removes the loop altogether, since it doesn't actually produce anything!

--
Mats

---

Code:

---


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


int main(int argc, char *argv[])
{
      clock_t start, end;
      double cpu_time_used;
 
      start = clock();

              int i = 0;
        int u = 0;
        int y = 0;
        int z = 0;

        for(i= 0;i++;i<5000000)
        {
                for(u= 0;u++;u<5000000)
                {
               
                        for(i= 0;i++;i<500000000)
                        {
                                z = z +1;
                                if(z > 1000000)
                                z = 0;       

                        }
               

                }

        }
       
      end = clock();
      cpu_time_used = ((double) (end - start)) / CLOCKS_PER_SEC;
        //cpu_time_used = ((double) (end - start)) /CLK_TCK;
        //printf("start: %e\n", start);
        //printf("end: %e\n",end);
          printf("end: %f\n",cpu_time_used);
}

---

cpu_time_used = 0.

ME ........ING ANGRY!

Try printing z after your loop!

--
Mats

just a fyi. your int main needs a return statement

So put a delay(500) inside the code you're timing, and knock 1/2 second off the final time it reports.

Quote:

---

Originally Posted by matsp

Try printing z after your loop!

--
Mats

---

Thanks.

I corrected it.

Quote:

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Originally Posted by CrustyClown

just a fyi. your int main needs a return statement

---

Thanks

Code:

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
















 int
      timeval_subtract (result, x, y)
          struct timeval *result, *x, *y;
      {
        /* Perform the carry for the later subtraction by updating Y. */
        if (x->tv_usec < y->tv_usec) {
          int nsec = (y->tv_usec - x->tv_usec) / 1000000 + 1;
          y->tv_usec -= 1000000 * nsec;
          y->tv_sec += nsec;
        }
        if (x->tv_usec - y->tv_usec > 1000000) {
          int nsec = (x->tv_usec - y->tv_usec) / 1000000;
          y->tv_usec += 1000000 * nsec;
          y->tv_sec -= nsec;
        }
 
        /* Compute the time remaining to wait.
          `tv_usec' is certainly positive. */
        result->tv_sec = x->tv_sec - y->tv_sec;
        result->tv_usec = x->tv_usec - y->tv_usec;
 
        /* Return 1 if result is negative. */
        return x->tv_sec < y->tv_sec;
      }























int main(int argc, char *argv[])
{
      clock_t start, end;
      double cpu_time_used;
        struct timespec echodelay;


      start = clock();
       
              int i = 0;
        int z = 0;
       

        for(i= 0;i<1000000000;i++)
        {
                z++;
                if (z > 1000)
                        z = 0;

        }
       
      end = clock();
      cpu_time_used = ((double) (end - start)) / CLOCKS_PER_SEC;
        //cpu_time_used = ((double) (end - start)) /CLK_TCK;
          printf("start: %e\n", start);
          printf("end: %e\n",end);


        timeval_subtract (&echodelay, &end, &start);
               
            int sec  = echodelay.tv_sec;
            int nsec = echodelay.tv_nsec;
           
          printf("Echo delay is %ds and %d ns\n", sec, echodelay.tv_nsec);
          printf("CPU: %f\n",cpu_time_used);
               
 
return 0;

}

---

At last I got it.

Echo delay is 4551208s and 478712 ns


This is 4.551208478712 sec?

You are printing rubbish. clock_t is not a timeval structure - clock_t is an integer (32-bit, most likely). Since you are passing the address of it, whatever comes AFTER your start/end variables will be taken as part of the timeval - and of course, that gives you some randomly curios value, not the actual difference between two clock() calls.

Also:

Code:

---

int
      timeval_subtract (result, x, y)
          struct timeval *result, *x, *y;

---

is OLD style C - newer (since 1989 at least) uses type declaration inside the parenthesis.

--
Mats

Rubbish...darn!

But it really took 4.5s.

Jebus!

I think the problem was your for loops:

Code:

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for(i= 0;i++;i<5000000)

---

which you fixed in your last version (you had the condition & increment statements mixed up).