C语言：输入一个人的出生年，月，日，再输入当前年，月，日，计算此人的实际年龄。求解各位C语大神。

Question

如题所述

Answer 1

　　要利用时间函数，代码如下：

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

int main()
{
    int year,month,day;
    printf("input birthday:\n");
    scanf("%d %d %d",&year,&month,&day);
    time_t cur = time(NULL);
    struct tm *today = localtime(&cur);//获得当前的日期
    int days0 = year*10000+(month-1)*100+day-1;
    int days1 = today->tm_year*10000+(today->tm_mon-1)*100+today->tm_mday-1;
    int age = days1-days0;//当日期或者月份不足的时候，会自动从上一级借位，因此计算出的年纪是可靠的，但是月份需要分析：需要减去99加12
    int mon = (age-(age/10000)*10000)/100;
    printf("age is %dyear %dmonth\n",age/10000,(mon>=12)?(mon-99+12);//计算出年龄
    return 0;
}

Answer 2

/*修改自http://www.jb51.net/article/39907.htm的程序*/
#include <stdio.h>
#define leapyear(year)((year) % 4 == 0 && ((year) % 100 != 0 || (year) % 400 == 0))

int days(int *day1, int *day2)
{
 int i=0;
 int *tmp;
 int diff = 0;
 const int month[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 if(day1[0] == day2[0])
 {
  if(day1[1] == day2[1])
  {
   diff = day1[2] - day2[2];
   diff = (diff < 0)?(-diff):diff;
  }
  else
  {
   if(day1[1] < day2[1])
   {
    tmp = day1;
    day1 = day2;
    day2 = tmp;
   }
   for(i=day2[1]+1; i<day1[1]; i++)
   {
    diff +=  month[i];
   }
   diff += month[day2[1]] - day2[2] + day1[2];
   if(day2[1] <= 2 && day1[1] >2)
    if(leapyear(day2[0]))
     diff++;
  }
 }
 else
 {
  if(day1[0] < day2[0])
  {
   tmp = day1; 
   day1 = day2;
   day2 = tmp;
  }
  for(i=day2[0]+1; i<day1[0]; i++)
  {
   if(leapyear(i))
    diff += 366;
   else
    diff += 365;
  }
  for(i=day2[1]+1; i<=12; i++)
  {
   diff += month[i];
  }
  diff += (month[day2[1]] - day2[2]);
  if(day2[1] <= 2)
   if(leapyear(day2[0]))
    diff++;
  for(i=1; i<day1[1]; i++)
  {
   diff += month[i];
  }
  diff += day1[2];
  if(day1[1] > 2)
   if(leapyear(day1[0]))
    diff++;
 }
 return diff;
}
void main(void)
{
 int day1[3], day2[3];
 int day = 0;

 printf("Input His/Her Birthday\n");
 scanf("%d-%d-%d",&day1[0], &day1[1], &day1[2]);
 printf("Input Current Date\n");
 scanf("%d-%d-%d",&day2[0], &day2[1], &day2[2]);
 day=days(day1, day2);
 printf("His/Her age is %2.2f\n" , (float)day/365);
 getch();
}

Answer 3

#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int age, byear, bmonth, bday, nyear, nmonth, nday;
printf("Please input the birthday(XXXX,XX,XX):");
scanf("%d,%d,%d", &byear, &bmonth, &bday);
printf("/nPlease input now date(XXXX,XX,XX):");
scanf("%d,%d,%d", &nyear, &nmonth, &nday);
age = nyear - byear;
if (nmonth > bmonth)
age++;
else if (nmonth == bmonth)
if (nday > bday)
age++;
printf("/nAge is %d/n", age);
}