Question

c语言两日期相差天数

#include <stdio.h>
#include<math.h>
void main()
{
int s=0,m1,d1,y1,y2,m2,d2,i,a[12]={31,28,31,30,31,30,31,31,30,31,30,31},b[12]={31,29,31,30,31,30,31,31,30,31,30,31};
scanf("%d %d %d",&y1,&m1,&d1);
scanf("%d %d %d",&y2,&m2,&d2);
if(y1==y2)
{
if(y1%100!=0&&y1%4==0||y1%400==0)
{
while(m1<m2)
{s=b[m1-1]+s;
m1=m1+1;
}
}
else
{
while(m1<m2)
{s=a[m1-1]+s;
m1=m1+1;
}
}
printf("%d days\n",s+d2-d1);
}
else
{
while(y1<y2)
{
if(m1<=2)
{
if(y1%100!=0&&y1%4==0||y1%400==0)
{s=366+s;}
else
{s=365+s;}
y1=y1+1;
}
else
{
y1=y1+1;
if(y1%100!=0&&y1%4==0||y1%400==0)
{s=366+s;}
else
{s=365+s;}
}
}
if(m1<=m2)
{
if(y2%100!=0&&y2%4==0||y2%400==0)
{for(i=0;m1<m2;i++)
{s=b[m1-1]+s;
m1=m1+1;
}
}
else
{for(i=0;m1<m2;i++)
{s=a[m1-1]+s;
m1=m1+1;
}
}
printf("%d days\n",s+d2-d1);
}
else
{
if(y2%100!=0&&y2%4==0||y2%400==0)
{
while(m1>m2)
{s=b[m1-1]+s;
m1=m1-1;
}
}
else
{
while(m1>m2)
{s=a[m1-1]+s;
m1=m1-1;
}
}
printf("%d days\n",s+d2-d1);
}
}
}

求高手指点错误所在
大体说下我的思路：
1.判断是否是同一年。
2.不同年，先计算相差的年数，再计算月份，最后计算天数

Answer 1

#include <stdio.h>
#include<math.h>
int main()
{
int s=0,m1,d1,y1,y2,m2,d2,i,a[12]={31,28,31,30,31,30,31,31,30,31,30,31},b[12]={31,29,31,30,31,30,31,31,30,31,30,31};
scanf("%d %d %d",&y1,&m1,&d1);
scanf("%d %d %d",&y2,&m2,&d2);
if(y1==y2)
{
if((y1%100!=0&&y1%4==0)||y1%400==0)
{
while(m1<m2)
{s=b[m1-1]+s;
m1=m1+1;
}
}
else
{
while(m1<m2)
{s=a[m1-1]+s;
m1=m1+1;
}
}
printf("%d days\n",s+d2-d1);
}
else
{
while(y1<y2)
{
if(m1<=2)
{
if((y1%100!=0&&y1%4==0)||y1%400==0)
{s=366+s;}
else
{s=365+s;}
y1=y1+1;
}
else
{
y1=y1+1;
if((y1%100!=0&&y1%4==0)||y1%400==0)
{s=366+s;}
else
{s=365+s;}
}
}
if(m1<=m2)
{
if((y2%100!=0&&y2%4==0)||y2%400==0)
{for(i=0;m1<m2;i++)
{s=b[m1-1]+s;
m1=m1+1;
}
}
else
{for(i=0;m1<m2;i++)
{s=a[m1-1]+s;
m1=m1+1;
}
}
printf("%d days\n",s+d2-d1);
}
else
{
if((y2%100!=0&&y2%4==0)||y2%400==0)
{
while(m1>m2)
{s=b[m1-1]+s;
m1=m1-1;
}
}
else
{
while(m1>m2)
{s=a[m1-1]+s;
m1=m1-1;
}
}
printf("%d days\n",s+d2-d1);
}
}
return 0;
}
你试试看！

追问

还是吧对。。你改哪了？

追答

if后面多加了一个括号！

Answer 2

说实话最好的方式是用时间戳的方式。在类unix系统中时间戳是从1970年1月1日00:00到某时间所经历的毫秒数，转化后两者相减再转化为天数就得出结果了。
"2014-04-09 22:00:02" => 2312312312321(到现在为止的时间长度)
"2013-12-11 12:33:12" => 231212311111
两者相差的毫秒数 => 转为天数