CODEWITHSUNDEEP
cgettimeofday
tshepang
tshepang

Reputation: 12459

Time from gettimeofday not updating as it should

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

int main()
{
   float time;
   struct timeval tv;
   gettimeofday( &tv, NULL );
   time = tv.tv_sec + ( tv.tv_usec / 1000000.0 );
   printf( "time: %f\n", time );
   return 0;
}

Running binary generated by this code repeatedly, I tend to get the same time value:

$ ./a.out
time: 1348059520.000000
$ ./a.out
time: 1348059520.000000
$ ./a.out
time: 1348059520.000000

This happens until several seconds later at which I get an updated time value.

Upvotes: 3

Views: 5410

Answers (2)

Jens
Jens

Reputation: 72746

Why do you use floating point at all?

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

int main (void)
{
   struct timeval tv;
   gettimeofday (&tv, NULL);
   printf ("time: %d.%06d\n", (int)tv.tv_sec, (int)tv.tv_usec);
   return 0;
}

./a.out
time: 1348067289.069908

It looks floaty, but it's inty :-) Since the microseconds value is in the range 0..999999 all you need to do is zero-pad it to 6 digits. No FP arithmetic required.

Upvotes: 3

tshepang
tshepang

Reputation: 12459

It appears that float is too small to contain the result of tv.tv_sec + ( tv.tv_usec / 1000000.0 ). Use double instead:

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

int main()
{
   double time;
   struct timeval tv;
   gettimeofday( &tv, NULL );
   time = tv.tv_sec + ( tv.tv_usec / 1000000.0 );
   printf( "time: %f\n", time );
   return 0;
}

Upvotes: 4

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