CODEWITHSUNDEEP
c
tomss
tomss

Reputation: 269

Measuring Execution Time

The result that is printed is always zero (0.00) and I want to know what I am doing wrong. The first function returns the beginning time and the second the final time. 

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

 void getTime(struct timeval begin){
    gettimeofday(&(begin), (struct timezone*) 0);
 }

 float elapTime(struct timeval begin, struct timeval end){
    return (1e+6*((end).tv_sec - (begin).tv_sec) + ((end).tv_usec - (begin).tv_usec));
 }

 int main(int argc, char **argv) {
     struct timeval begin, end;
     getTime(begin);

     printf("Time: %.2f", elapTime(begin, end));
 }

Upvotes: 2

Views: 116

Answers (2)

hawkeye
hawkeye

Reputation: 169

You could do something like this instead:

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

int main() {
  clock_t begin, end;
  int i = 1e8;

  begin = clock();
  while(i--);
  end = clock();

  printf("Time: %.2f\n", (double) (end-begin)/CLOCKS_PER_SEC);
}

The clock() function counts processor clock cycles. By dividing it by CLOCKS_PER_SEC you get seconds. The code above counts the time it takes to iterate 1e8 down to 0.

Upvotes: 1

lain
lain

Reputation: 448

Try using more simple functions:

double              floattime (void)
{
    struct timeval  t;

    if (gettimeofday (&t, (struct timezone *)NULL) == 0)
        return (double)t.tv_sec + t.tv_usec * 0.000001;
    return (0);
}

int main(int argc, char **argv) {
    double begin;
    begin = floattime();

    getchar ();
    printf("Time: %.2f", floattime () - begin);
    return 0;
 }

And don't forget to wait some time before calculating time execution. Else, it will always return 0.00s.

Upvotes: 0

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