有的时候我会想,再过个100天是几月几日呢?也许你可以翻翻日历,但是如果是1000天、10000天以后呢?为什么我们不写个小程序来专门处理这个问题呢?似乎也不是很复杂……
于是这篇博客应运而生,其实计算的逻辑很简单,就是使用计算机模拟日期计数,比较有特色的地方在于switch-case的“非常规”运用——
/ *
* 文件名:nextdate.c
* 作者:于子豪
* 描述:计算多少天后的日期
* 日期:2019-7-30
*/
#include <stdio.h>
typedef struct date
{
int year;
int month;
int day;
}Date;
//是否是闰年判断
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)<-28) {day-=28;today.month++;}
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;
}
}
}
//测试程序
int main(int argc, char *argv[])
{
Date today={1999,1,11};
int days=117+365;
printf("今天是%d.%d.%d:经过%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;
}
最后,附上一个C-Free5的压缩包:
https://download.csdn.net/download/qq_45467083/11453165