How do I compare two timestamps in C?

Question

I\'m writing a socket program that maintains FIFO queues for two input sockets. When deciding which queue to service, the program pulls the most recent time-stamp from each

Answer 1

timercmp() is just a macro in libc (sys/time.h):

# define timercmp(a, b, CMP)                                                  \
  (((a)->tv_sec == (b)->tv_sec) ?                                             \
   ((a)->tv_usec CMP (b)->tv_usec) :                                          \
   ((a)->tv_sec CMP (b)->tv_sec))

If you need timersub():

# define timersub(a, b, result)                                               \
  do {                                                                        \
    (result)->tv_sec = (a)->tv_sec - (b)->tv_sec;                             \
    (result)->tv_usec = (a)->tv_usec - (b)->tv_usec;                          \
    if ((result)->tv_usec < 0) {                                              \
      --(result)->tv_sec;                                                     \
      (result)->tv_usec += 1000000;                                           \
    }                                                                         \
  } while (0)

Answer 2

googling timeval give this first result. From that page:

It is often necessary to subtract two values of type struct timeval or struct timespec. Here is the best way to do this. It works even on some peculiar operating systems where the tv_sec member has an unsigned type.

 /* Subtract the `struct timeval' values X and Y,
    storing the result in RESULT.
    Return 1 if the difference is negative, otherwise 0.  */

 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;
 }

Answer 3

For viewing timevals I just whipped this up. It returns a timeval as a string that you can print or send to a text file:

char *tv2str(struct timeval *intv) {
  static char ans[200];
  snprintf(ans,200,"%u.%u",(unsigned int)intv->tv_sec, \
    (unsigned int) intv->tv_usec);
  return ans;
}

Use like:

printf("nowtv: %s\n",tv2str(&nowtv));

nowtv: 1568407554.646623

Timercmp() didn't seem to work right so I wanted a way to check up on it by actually looking at some values.

Answer 4

This is slightly different, but I think clearly illustrates the logic involved. I'm working on some MSP430 code in C, and have a timestamp struct very similar to timeval, but with nsecs instead of usecs.

This code keeps everything positive, so unsigned ints would work fine, and avoids overflows (I think). It also doesn't modify the timestamps/timevals being passed in, except the result of course.

typedef struct timestamp {
    int32_t secs;
    int32_t nsecs;
} timestamp_t;

int timestamp_sub(timestamp_t * x, timestamp_t * y, timestamp_t * result){
    // returns 1 if difference is negative, 0 otherwise
    // result is the absolute value of the difference between x and y
    negative = 0;
    if( x->secs > y->secs ){
        if( x->nsecs > y->nsecs ){
            result->secs = x->secs - y->secs;
            result->nsecs = x->nsecs - y->nsecs;
        }else{
            result->secs = x->secs - y->secs - 1;
            result->nsecs = (1000*1000*1000) - y->nsecs + x->nsecs;
        }
    }else{
        if( x->secs == y->secs ){
            result->secs = 0;
            if( x->nsecs > y->nsecs ){
                result->nsecs = x->nsecs - y->nsecs;
            }else{
                negative = 1;
                result->nsecs = y->nsecs - x->nsecs;
            }
        }else{
            negative = 1;
            if( x->nsecs > y->nsecs ){
                result->secs = y->secs - x->secs - 1;
                result->nsecs = (1000*1000*1000) - x->nsecs + y->nsecs;
            }else{
                result->secs = y->secs - x->secs;
                result->nsecs = y->nsecs - x->nsecs;
            }
        }
    }
    return negative;
}