Question

我正在尝试编写一个c程序，用于确定用户提供日期的一年中的日期编号（1到366）。我如何调整这个程序闰年，如（2012年12月31日）是第366天，因为它是闰年。到目前为止这是我的代码。我想我在切换之前需要一个if staement但是我不知道如何实现请帮助

#include <stdio.h>

int isleap(int year);

void  displaydate(int month ,int date ,int year,int daynumber);

int main (void) {

    int month;
    int day;
    int year;
    int leapyear;
    int daynumber;

    printf(" enter a month");
    scanf("%d",&month);

    printf(" enter a day");
    scanf("%d",&day);

    printf(" enter a year");
    scanf("%d",&year);

    leapyear= isleap( year);

    displaydate( month,day,year,daynumber);

    daynumber = leapyear && month > 2 ? day+1 : day;

    switch(month) {

        case 1 : daynumber = day;
                break;

        case 2 : daynumber = 31 + day;
                break;

        case 3 : daynumber = 59 + day + 1; 
                break;

        case 4 : daynumber = 89 + day;
                break;  

        case 5 : daynumber = 120 + day;
                break;  

        case 6 : daynumber = 150 + day;
                break; 

        case 7 : daynumber = 181 + day;
                break;  

        case 8 : daynumber = 212 + day;
                break;   

        case 9 : daynumber = 232 + day;
                break;  

        case 10 : daynumber = 263 + day;
                break;  

        case 11 : daynumber = 293 + day;
                break;  

        case 12 : daynumber = 324 + day;
                break;

        default:
                break;
    }

    return 0;
}

int isleap(int year)
{
    if (((year % 4 == 0) && (year % 100!= 0)) || (year%400 == 0)) {
        printf("%d  1", year);
        return 1;
    }
    else {
        printf("%d 0", year);
        return 0; 
    }
}

void  displaydate(int month ,int date ,int year,int daynumber)
{ 
    printf("\n your date is %d and the day number is %d",month,date,year,daynumber);

}

Answer 1

您可以使用三元运算符，只需在切换之前添加此代码：

day = leapyear && month > 2 ? day+1 : day;

Answer 2

int isleap(int year)
{
    if (((year % 4 == 0) && (year % 100!= 0)) || (year%400 == 0))
        printf("%d  1", year);
    else
        printf("%d 0", year);

}

上面的函数应该返回是否是闰年，所以更改代码以正确返回如下。

int isleap(int year)
{
    if (((year % 4 == 0) && (year % 100!= 0)) || (year%400 == 0)){
        printf("%d  1", year);
        return 1;
    }else{
        printf("%d 0", year);
        return 0; }

}

现在在上面添加以下语句切换案例，如果它是闰年且月份大于2，则增加一天。

day = leapyear  && (month>2) ? day+1 : day;

现在你将得到正确的结果。

希望这会有所帮助。

Answer 3

除了提出的其他问题之外，您在switch声明中添加的天数也是错误的。它应该是：

    switch (month) {
        case 1 :  daynumber = day; break;
        case 2 :  daynumber = 31  + day + leapyear; break;
        case 3 :  daynumber = 59  + day + leapyear; break;
        case 4 :  daynumber = 90  + day + leapyear; break;  
        case 5 :  daynumber = 120 + day + leapyear; break;  
        case 6 :  daynumber = 151 + day + leapyear; break; 
        case 7 :  daynumber = 181 + day + leapyear; break;  
        case 8 :  daynumber = 212 + day + leapyear; break;   
        case 9 :  daynumber = 243 + day + leapyear; break;  
        case 10 : daynumber = 273 + day + leapyear; break;  
        case 11 : daynumber = 304 + day + leapyear; break;  
        case 12 : daynumber = 334 + day + leapyear; break;
        default: break;
    }

您在计算之前也无法显示daynumber，您需要按如下方式对计算进行排序：

    ...
    leapyear = isleap (year);

    switch (month) {
        case 1 :  daynumber = day; break;
        case 2 :  daynumber = 31  + day + leapyear; break;
        case 3 :  daynumber = 59  + day + leapyear; break;
        case 4 :  daynumber = 90  + day + leapyear; break;  
        case 5 :  daynumber = 120 + day + leapyear; break;  
        case 6 :  daynumber = 151 + day + leapyear; break; 
        case 7 :  daynumber = 181 + day + leapyear; break;  
        case 8 :  daynumber = 212 + day + leapyear; break;   
        case 9 :  daynumber = 243 + day + leapyear; break;  
        case 10 : daynumber = 273 + day + leapyear; break;  
        case 11 : daynumber = 304 + day + leapyear; break;  
        case 12 : daynumber = 334 + day + leapyear; break;
        default: break;
    }

    displaydate (month, day, year, daynumber);
    ...

printf中的displaydate语句无法为您传递的每个参数提供转换说明符。看来你打算：

void displaydate (int month, int date, int year, int daynumber)
{ 
    printf ("\n your date is %d/%d/%d and the day number is %d\n",
            month, date, year, daynumber);
}

注意： "%d/%d/%d"

的mm/dd/yyyy

如果你将所有这些部分放在一起，你会得到：

#include <stdio.h>

int isleap (int year);

void displaydate (int month, int date , int year, int daynumber);

int main (void) {

    int month = 0,
        day = 0,
        year = 0,
        leapyear = 0,
        daynumber = 0;

    printf (" enter month: ");
    scanf ("%d", &month);

    printf (" enter day: ");
    scanf ("%d", &day);

    printf (" enter year: ");
    scanf ("%d", &year);

    leapyear = isleap (year);

    switch (month) {
        case 1 :  daynumber = day; break;
        case 2 :  daynumber = 31  + day + leapyear; break;
        case 3 :  daynumber = 59  + day + leapyear; break;
        case 4 :  daynumber = 90  + day + leapyear; break;  
        case 5 :  daynumber = 120 + day + leapyear; break;  
        case 6 :  daynumber = 151 + day + leapyear; break; 
        case 7 :  daynumber = 181 + day + leapyear; break;  
        case 8 :  daynumber = 212 + day + leapyear; break;   
        case 9 :  daynumber = 243 + day + leapyear; break;  
        case 10 : daynumber = 273 + day + leapyear; break;  
        case 11 : daynumber = 304 + day + leapyear; break;  
        case 12 : daynumber = 334 + day + leapyear; break;
        default: break;
    }

    displaydate (month, day, year, daynumber);

    return 0;
}

int isleap (int year)
{
    if (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0))
        return 1;

    return 0; 
}

void displaydate (int month, int date, int year, int daynumber)
{ 
    printf ("\n your date is %d/%d/%d and the day number is %d\n",
            month, date, year, daynumber);
}

此外，请注意，您必须验证所有用户输入（上面省略）。至少应该检查每个scanf的回报，类似于：

    if (scanf ("%d", &month) != 1) {
        fprintf (stderr, "error: invalid month entered.\n");
        return 1;
    }

仔细看看，如果您有任何问题，请告诉我。

确定日期编号提示的程序

3 个答案: