C語言:計算多少天后的日期

有的時候我會想，再過個100天是幾月幾日呢？也許你能夠翻翻日歷，可是若是是1000天、10000天之后呢？為何咱們不寫個小程序來專門處理這個問題呢？彷佛也不是很復雜……

因而這篇博客應運而生，其實計算的邏輯很簡單，就是使用計算機模擬日期計數，比較有特點的地方在于

switch-case的“很是規”運用——

/ *

* 文件名：nextdate.c

* 做者：于子豪

* 描述：計算多少天后的日期

* 日期：2019-7-30

*/

#include

typedef struct date

{

int year;

int month;

int day;

}Date;

//是不是閏年判斷

int isbissextile(int year)

{

if(((year%100!=0)&&(year%4==0))||year%400==0)

return 1;

else return -1;

}

Date todate(Date today, int days)

{

int day=today.day+days;

while(1)

{

switch(today.month)

{

case 1:

if(day>31) {day-=31;today.month++;}

else {today.day=day;return today;}

case 2:

if(day*isbissextile(today.year)>29) {day-=29;today.month++;}

else if(day*isbissextile(today.year)

else {today.day=day;return today;}

case 3:

if(day>31) {day-=31;today.month++;}

else {today.day=day;return today;}

case 4:

if(day>30) {day-=30;today.month++;}

else {today.day=day;return today;}

case 5:

if(day>31) {day-=31;today.month++;}

else {today.day=day;return today;}

case 6:

if(day>30) {day-=30;today.month++;}

else {today.day=day;return today;}

case 7:

if(day>31) {day-=31;today.month++;}

else {today.day=day;return today;}

case 8:

if(day>31) {day-=31;today.month++;}

else {today.day=day;return today;}

case 9:

if(day>30) {day-=30;today.month++;}

else {today.day=day;return today;}

case 10:

if(day>31) {day-=31;today.month++;}

else {today.day=day;return today;}

case 11:

if(day>30) {day-=30;today.month++;}

else {today.day=day;return today;}

case 12:

if(day>31) {day-=31;today.month=1;today.year++;}

else {today.day=day;return today;}

default:break;

}

}

}

//測試程序

int main(int argc, char *argv[])

{

Date today={1999,1,11};

int days=117+365;

printf("今天是%d.%d.%d：通過%d天后是：",today.year,today.month,today.day,days);

Date nextday=todate(today,days);

printf("%d.%d.%d\n",nextday.year,nextday.month,nextday.day);

return 0;

}

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