[求助]：天数与日期的转换问题，小弟新手！

*/ --------------------------------------------------------------------------------------
*/ 出自： 编程中国 http://www.bc-cn.net
*/ 作者： 李飞龙 QQ：81516156
*/ 时间： 2007-9-18 编程论坛首发
*/ 声明： 尊重作者劳动，转载请保留本段文字
*/ --------------------------------------------------------------------------------------
main()
{static int day_tab[][13]={
{0,31,28,31,30,31,30,31,31,30,31,30,31},
{0.31,29,31,30,31,30,31,31,30,31,30,31}};
printf("%d\n",day_of_year(day_tab,2007,100));
}
day_of_year(day_tab,year,yearday,pmonth,pday)
int year,yearday,*pmonth,*pday;
{
int i,leap,q;
leap=(year%4==0&&year%100||year%400==0);
for(i=1;yearday>(q=*(*(day_tab+leap)+i));i++)
yearday-=q;
*pmonth=i;
*pday=yearday;
}

实在不知道哪儿错了？
望大侠指导一下！！！

----------------解决方案--------------------------------------------------------

void main()
{
static int day_tab[][13]={
{0,31,28,31,30,31,30,31,31,30,31,30,31},
{0.31,29,31,30,31,30,31,31,30,31,30,31}};
printf("%d\n",day_of_year(day_tab,2007,100));///////////
}

int day_of_year(day_tab,year,yearday,pmonth,pday)////////////
int year,yearday,*pmonth,*pday;
{
int i,leap,q;
leap=(year%4==0&&year%100||year%400==0);
for(i=1;yearday>(q=*(*(day_tab+leap)+i));i++)
yearday-=q;
*pmonth=i;
*pday=yearday;
}

你的函数调用的都不正确啊。还有就是你以后写程序请注意自己程序的清晰。

----------------解决方案--------------------------------------------------------

请问这些东西
void main()
{
static int day_tab[][13]={
{0,31,28,31,30,31,30,31,31,30,31,30,31},
{0.31,29,31,30,31,30,31,31,30,31,30,31}};
printf("%d\n",day_of_year(day_tab,2007,100));///////////
}

int day_of_year(day_tab,year,yearday,pmonth,pday)////////////
int year,yearday,*pmonth,*pday;
{
int i,leap,q;
leap=(year%4==0&&year%100||year%400==0);
for(i=1;yearday>(q=*(*(day_tab+leap)+i));i++)
yearday-=q;
*pmonth=i;
*pday=yearday;
都是要用手写打出来吗?
小弟新手

----------------解决方案--------------------------------------------------------

的确这格式看都看不懂，应该改正确了在发上来。{}匹配都不对嘛
----------------解决方案--------------------------------------------------------

楼主错误好多，在主函数中应声明两个指针，分别存放月份和当月的天数，函数返回时得到相应的月数和天数
主函数：
static int day_tab[][13]={
{0,31,28,31,30,31,30,31,31,30,31,30,31},
{0,31,29,31,30,31,30,31,31,30,31,30,31}};
int main()
{

int *p_month,*p_day;
void day_of_year(int,int,int *,int *);
p_month=new (int);
p_day=new (int);
day_of_year(2007,100,p_month,p_day);
printf("%d月 %d日\n",*p_month,*p_day);
delete p_day;
delete p_month;
return 0;
}
void day_of_year(int year,int yearday,int *pmonth,int *pday)
{
int i,leap,q;
leap=(year%4==0&&year%100||year%400==0);
for(i=1;yearday>(q=*(*(day_tab+leap)+i));i++)
yearday-=q;
*pmonth=i;
*pday=yearday;
}

----------------解决方案--------------------------------------------------------