我想计算两个日期之间的秒数差异,但结果是0。
以下是代码:
time_t=time(NULL);
struct tm * timeNow=localtime();
time_t start=mktime(&*timeNow);
time_t end=mktime(&*recordFind->timeInserted);
double seconds=difftime(start,end);
recordFind->timeInserted
没问题因为我打印了他的成员并且没事,
但是当我打印秒数是0.000000;
答案 0 :(得分:5)
你想要
double seconds = difftime(end, start);
而不是
double seconds = difftime(start, end);
并且您忘记将变量命名为time_t=time(NULL);
,更改为:
time_t now;
double seconds;
time(&now);
seconds = difftime(now, mktime(&recordFind->timeInserted));
答案 1 :(得分:1)
请参阅下文
#include <stdio.h>
struct TIME
{
int seconds;
int minutes;
int hours;
};
void differenceBetweenTimePeriod(struct TIME t1, struct TIME t2, struct TIME *diff);
int main()
{
struct TIME startTime, stopTime, diff;
printf("Enter start time: \n");
printf("Enter hours, minutes and seconds respectively: ");
scanf("%d %d %d", &startTime.hours, &startTime.minutes, &startTime.seconds);
printf("Enter stop time: \n");
printf("Enter hours, minutes and seconds respectively: ");
scanf("%d %d %d", &stopTime.hours, &stopTime.minutes, &stopTime.seconds);
// Calculate the difference between the start and stop time period.
differenceBetweenTimePeriod(startTime, stopTime, &diff);
printf("\nTIME DIFFERENCE: %d:%d:%d - ", startTime.hours, startTime.minutes, startTime.seconds);
printf("%d:%d:%d ", stopTime.hours, stopTime.minutes, stopTime.seconds);
printf("= %d:%d:%d\n", diff.hours, diff.minutes, diff.seconds);
return 0;
}
void differenceBetweenTimePeriod(struct TIME start, struct TIME stop, struct TIME *diff)
{
if(stop.seconds > start.seconds){
--start.minutes;
start.seconds += 60;
}
diff->seconds = start.seconds - stop.seconds;
if(stop.minutes > start.minutes){
--start.hours;
start.minutes += 60;
}
diff->minutes = start.minutes - stop.minutes;
diff->hours = start.hours - stop.hours;
}
输出
Enter start time:
Enter hours, minutes and seconds respectively:
12
34
55
Enter stop time:
Enter hours, minutes and seconds respectively:
8
12
15
TIME DIFFERENCE: 12:34:55 - 8:12:15 = 4:22:40