问题描述:

What's the simplest way to calculate the number of minutes until 9pm (being today or tomorrow)?

Was looking over mktime() but seemed a hassle to get the day # and month, dealing with all that.

网友答案:

Just calculate the delta to today's 9pm. If it's negative, add 24hrs to get the delta to tomorrow's 9pm.

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

int main(int argc, char* argv[]) {
    struct tm* tm;
    time_t ts=time(NULL);
    if(tm=localtime(&ts)) {
            long int delta;
            tm->tm_hour = 21;
            tm->tm_min = 0;
            tm->tm_sec = 0;
            delta = mktime(tm) - ts;
            if(delta<0) {
                    delta+=24*60*60;
            }
            printf("%d minutes til (next)9pm\n",delta/60);
    }
    return 0;
}
网友答案:

Here it is (output in seconds...)

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

int main(void) 
{
  double dt = 0.0f;
  time_t ti = time(NULL);
  time_t tf;
  struct tm *pt_actual = localtime(&ti);

  struct tm t_actual = *pt_actual;
  struct tm t_target = *pt_actual;

  t_target.tm_hour = 21; // 09:00 pm
  t_target.tm_min = 0;

  tf = mktime(&t_target);

  dt = difftime(tf, ti);
  if (dt < 0)
    dt += (24*60*60);  // if negative, just consider **next** 9pm target, add a day

  printf("difftime = %f seconds\n", dt );
  return 0;
}
网友答案:

This will give you the minutes until 9pm.:

time_t currenttime;
time_t ninePM;
struct tm * timeinfo;
int minutesUntilNine;

// Fill currenttime + convert to struct tm
time ( &currenttime );
timeinfo = localtime ( &rawtime );

// Check for after 9pm...
if (timeinfo->tm_hour >= 21) // Are we after 9pm?
  timeinfo->tm_day++; // increment our day

// Set timeinfo to the time you want...
timeinfo->tm_sec = 0;
timeinfo->tm_min = 0;
timeinfo->tm_hour = 21; // 9pm
ninePM = mktime( timeinfo );

minutesUntilNine = (ninePM - currenttime) / 60;
网友答案:

A simple way is to calculate the number of hours and minutes (given what the time is now, plus the fact that there are 24 hours in a day), and multiply the number of hours by 60 to make it a number of minutes. Alternatively, instead of using a 24-hour clock to express the time, use a 1440 minute clock: it's easy to subtract one time from another, to get a time difference.

网友答案:

The guy's asking for the simplest way. All your ways are unnecessarily complex. Just use modular arithmetic:

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

#define TARGET_TIME 21*60 // 9:00 PM, in format HH * 60 + MM

int main(int argc, char* argv[])
{
    struct tm* tm;
    time_t ts = time(NULL);
    int minutes_until_9pm;
    tm = localtime(&ts);
    minutes_until_9pm = 24*60 - (tm->tm_min + tm->tm_hour * 60 + TARGET_TIME)%(24*60);

    //alternatively, the following is slightly faster
    //minutes_until_9pm = 24*60 - (tm->tm_min + tm->tm_hour * 60 + TARGET_TIME) + (tm->tm_min+tm->tm_hour*60+TARGET_TIME>24*60?24*60:0);

    printf("%d minutes til (next)9pm\n", minutes_until_9pm);
    return 0;
}

If you know your timezone , it all becomes even simpler:

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

#define TARGET_TIME 21*60 // 9:00 PM, in format HH * 60 + MM
#define TIMEZONE (-6) // This is CST; replace with your own timezone

int main(int argc, char* argv[])
{
    int minutes_until_9pm = 24*60 - (time(NULL)/60 + TARGET_TIME + (24+TIMEZONE)*60)%(24*60);
    printf("%d minutes til (next)9pm\n", minutes_until_9pm);
    return 0;
}

What's going on here is that, modulo one day, we're getting number of minutes since midnight. Then we're adding 9:00 PM to get the the number of minutes since 9:00 PM. Then we're subtracting that from 24 hours to get the number of minutes until the next 9:00 PM.

If you need higher accuracy (i.e. rounded to the nearest number of minutes, rather than simply number of minutes), simply subtract (tm->tm_sec>=30?0:1) from minutes_until_9pm, or replace time(NULL)/60 with (time(NULL)+30)/60, respectively.

网友答案:

C99 version:

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

int main(void)
{
    struct tm now = *localtime(&(time_t){ time(NULL) });
    struct tm then = now;
    then.tm_hour = 21;
    then.tm_min = 0;
    then.tm_sec = 0;
    int dt = ((int)round(
        difftime(mktime(&then), mktime(&now)) / 60) + 24 * 60) % (24 * 60);
    printf("time till 9pm: %i minutes\n", dt);
}
网友答案:

The more profound, if less helpful, answer is that there is no simple answer.

How do you even know that today will have a 9pm? OK, sure, I'm sure all the time zones that exist never skip 21:00, but they skip hours in the middle of the night once a year.

Calendrics is a disaster of complexity. Beware.

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