Question

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

int date_txt[5],today_date;

scanf("%d",&today_date);        //i enter (input) the date manually e.g. 20171109
date_txt[0]=20161102;           // year month day form (rrrrmmdd)
date_txt[1]=20150101;
date_txt[2]=20170615;
date_txt[3]=20160628;
date_txt[4]=20150101;

我有几个日期需要与today_date进行比较，看看差异是否等于或大于1年。我发现存在一个叫做difftime（）的函数，但由于我是新手，我不知道如何做到这一点。感谢任何帮助，谢谢。

Answer 1

您可以这样做：

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

 #define COUNT 5
 #define SECONDS_NON_LEAP_YEAR (365*24*3600)
 #define SECONDS_LEAP_YEAR (366*24*3600)

 typedef struct tm T_TIME;
 T_TIME dToday, dTemp;

 void convertToTime(int iDate, T_TIME* tTime)
 {
    tTime->tm_mday = iDate%100;
    iDate/=100;
    tTime->tm_mon = (iDate%100)-1;
    iDate/=100;
    tTime->tm_year = iDate-1900;
 }

 int isLeapYear(int yyyy)
 {
    if(yyyy%100) // Not a century
    {
        return !(yyyy%4);
    }
    return !(yyyy%400);
 }

 int main(void) {
    int date_txt[]={20161102, 20150101, 20170615, 20160628, 20150101}; 
    int today_date, i, flg, isThisYearLeap, tSeconds;

    scanf("%d",&today_date); //i enter (input) the date manually e.g. 20171109 year month day form (rrrrmmdd)
    convertToTime(today_date, &dToday);
    isThisYearLeap = isLeapYear(dToday.tm_year+1900);

    for(i=0;i<COUNT; i++)
    {
        convertToTime(date_txt[i], &dTemp);
        tSeconds = difftime(mktime(&dToday), mktime(&dTemp));
        flg = (isThisYearLeap || isLeapYear(dTemp.tm_year+1900)) ?
            (tSeconds >= SECONDS_LEAP_YEAR) : (tSeconds >= SECONDS_NON_LEAP_YEAR);

        printf("The difference between Today and date %d is %s 1 year.\n", date_txt[i],
                    (flg ? "Greater than or equals to" : "Less than"));
    }

    return 0;
 }

在执行here

中查看

如何使用difftime（）比较日期？

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