C语言计算日期间隔天数

要计算两个日期相关多少天，

网上看到一个“C语言计算日期间隔天数的经典算法解析”文章，其代码如下：

int day_diff(int year_start, int month_start, int day_start, int year_end, int month_end, int day_end)
{
	int y2, m2, d2;
	int y1, m1, d1;
	
	m1 = (month_start + 9) % 12;
	y1 = year_start - m1/10;
	d1 = 365*y1 + y1/4 - y1/100 + y1/400 + (m1*306 + 5)/10 + (day_start - 1);
 
	m2 = (month_end + 9) % 12;
	y2 = year_end - m2/10;
	d2 = 365*y2 + y2/4 - y2/100 + y2/400 + (m2*306 + 5)/10 + (day_end - 1);
	
	return (d2 - d1);
}

我用java随机生成一些数据去验证这个c代码，有些数据不对，比如

4645-8-25
5074-4-28

java计算如下：

(int) ((c2.getTime().getTime() - c1.getTime().getTime()) / (1000 * 3600 * 24));

然后自己写了个代码，

u8 IsLeapYear(u16 year){
	if ((year & 0x03) == 0) {
		if ((year >> 2) % 25 == 0) {
			if ((year & 0x0f) == 0)
				return 1;
		} else {
			return 1;
		}
	}
	
	return 0;
}
const u8 offset[]= {0,0,1,5,6,8,9,11,12,13,15,16,18};
u32 CalDayDiff(MyDate* s,MyDate* e){
	u32 yDiff,i;
	u32 sum=0,sum2;
	MyDate d;
	
	yDiff=e->year-s->year;
	
	if(yDiff>1){
		for(i=s->year+1;i<e->year;i++){
			sum+=IsLeapYear(i)?366:365;
		}
		yDiff=1;
	}
	
	switch(yDiff){
		case 0:
			sum=((e->month-1)<<5)-offset[e->month]+e->day;
			if(e->month>2 && IsLeapYear(e->year))
				sum+=1;
			sum2=((s->month-1)<<5)-offset[s->month]+s->day;
			if(s->month>2 && IsLeapYear(s->year))
				sum2+=1;
				
			sum-=sum2;
			break;
		case 1:
			d.year=s->year;
			d.month=12;
			d.day=31;
			sum+=CalDayDiff(s,&d);
			d.year=e->year;
			d.month=1;
			d.day=1;
			sum+=CalDayDiff(&d,e)+1;
			break;
	}
	
	return sum;
}