Question

我的C编码越来越好，但我仍然不确定如何在我的代码中解决几个问题。教授希望我们写一个输入日期的代码（mm / dd / yyyy）并吐出已经过去的天数。对于选择2，输入特定年份的天数并吐出相应的日期。问题是（因为我们的教授喜欢弄乱我们）解决像4/31/2001这样的错误，其中4月没有31天，并且在非闰年输入了366天。

int leap(int x)
{
if ((x % 4 == 0 && x % 100 != 0)|| x % 400 == 0)
    return 1;
else
    return 0;
}


 int monthFun(int d, int y)
{
int days[] = {31,28,31,30,31,30,31,31,30,31,30,31};
int i;


if (leap(y))
    days[1] = 29;

for (i = 0; ; i++)
    if (d > days[i])
        d = d - days[i];
    else
        return i + 1;
return i + 1;


}


int dateFun(int d, int y)
{
int days[] = {31,28,31,30,31,30,31,31,30,31,30,31};
int i, m;

if (leap(y))
    days[1] = 29;
for (i = 0; i < m - 1; i++)
    if (d > days[i]) {
       d = d - days[i];
    }
    else {
        d = d;
        m = i;

    }

return d;



}

int daysPast(int m, int d, int y)
{
int days[] = {31,28,31,30,31,30,31,31,30,31,30,31};
int i, past = 0;

if (leap(y))
    days[1] = 29;

for (i = 0; i < m - 1; i++)
    past = past + days[i];
return past + d;
}



int main()
{
int mm, dd, yy, choice;
char more;

do {
    printf ("\n\t\tThis program will find days passed or date in the year");
    printf ("\n\t\t\t\t1) Input date (mm/dd/yyyy) to find the days passed");
    printf ("\n\t\t\t\t2) Input passed days to find date in the year");
    printf ("\n\n\t\t\t\tYour choice (1/2): ");
    scanf  ("%d", &choice);

    if (choice == 1)    {
        printf ("\n\t\tPlease input date (mm/dd/yyyy): ");
        scanf  ("%d/%d/%d", &mm, &dd, &yy);
        printf ("\n\t\tThere are %d days passed in the year %d", daysPast(mm, dd, yy), yy);
    }
    else
        if (choice == 2){
            printf ("\n\t\tInput days: ");
            scanf  ("%d", &dd);
            printf ("\n\t\tInput year: ");
            scanf  ("%d", &yy);
            printf ("\n\t\tThe date is %d/%d/%d", monthFun(dd, yy), dateFun(dd,yy), yy);

        }

        else
            printf ("\n\t\tPlease choose 1 or 2");
    printf ("\n\n\t\tDo more (Y/N)? ");
    scanf  ("%s", &more);
} while (more == 'y' || more == 'Y');
}

我的dateFun（）的代码有点草率（或者至少对我而言）。我想如果输入上述问题之一，程序会说“无效输入”之类的东西，并带你回到“做更多”。任何提示都表示赞赏，并提前感谢！

Answer 1

你的dateFun测试一个未初始化的m。

dateFun（）和monthFun（）看起来非常相似，我添加了一个示例组合版本。

int leap(int x)
{
  return ((x % 4 == 0 && x % 100 != 0)|| x % 400 == 0);
}


int monthFun(int d, int y)
{
  int i, days[] = {31,28,31,30,31,30,31,31,30,31,30,31};

  days[1] += leap(y);

  for (i=0 ; d>days[i] ; i++)
    d -= days[i];
  return i+1;
}


int dateFun(int d, int y)
{
  int i, days[] = {31,28,31,30,31,30,31,31,30,31,30,31};

  days[1] += leap(y);

  for (i=0 ; d>days[i] ; i++)
    d -= days[i];
  return d;
}

int datemonthFun(int d, int y)
{
  int i, days[] = {31,28,31,30,31,30,31,31,30,31,30,31};

  days[1] += leap(y);

  for (i=0 ; d>days[i] ; i++)
    d -= days[i];
  return (i+1)*100+d;
}

解决通过函数传递的日期和天数

1 个答案: