c - C语言的日期计算器程序

Question

我正在尝试用 C 程序编写一个范围从 1/1/1902 到 12/31/2299 的日期计算器，我遵循了 http://en.wikipedia.org/wiki/Calculating_the_day_of_the_week 中的算法有世纪表、月表和日表，但是当我尝试打印出来时，这就是我所拥有的

Enter Year, Month and Day as YYYY,M,DD
1982 4 24
Saturday1982 ,6, 24, is  6

而不是说 1982 4 24 是星期六

我写的程序有什么问题？开关盒的位​​置？

#include <stdio.h>

int main (void)
{
    // insert code here...
    int day,month,year;

    printf("Enter Year, Month and Day as YYYY,M,DD\n");

    scanf("%4d%d%2d", &year, &month, &day);

    if (year >= 1901 && year <= 2299 &&
        month >= 1 && month <= 12 &&
        day >= 0 && day <= 31)
    {
        int century = year/100;

        /* making a century table for calculation*/

        switch(century)
        {
            case 19:
                century=0;
                break;
            case 20:
                century=6;
                break;
            case 21:
                century=4;
                break;
            case 22:
                century=2;
                break;
        }

        int last2_of_year= year % 100;

        /* Last 2 digits of the year entered*/

        int last2_div_4 =  last2_of_year/4;

        switch (month)
        {
            case 1:
                month=0;
                break;

            case 2:
                month=3;
                break;
            case 3:
                month=3;
                break;
            case 4:
                month=6;
                break;
            case 5:
                month=1;
                break;
            case 6:
                month=4;
                break;
            case 7:
                month=6;
                break;
            case 8:
                month=2;
                break;
            case 9:
                month=5;
                break;
            case 10:
                month=0;
                break;
            case 11:
                month=3;
                break;
            case 12:
                month=5;
                break;
         }


        int total_num = (century+ last2_of_year +day +month +last2_div_4)%7;

        switch (total_num) 
        {
            case 0:
            printf("Sunday");
                break;
            case 1:
            printf("Monday");
                break;
            case 2:
            printf("Tuesday");
                break;
            case 3:
            printf("Wednesday");
                break;
            case 4:
            printf("Thursday");
                break;
            case 5:
            printf("Friday");
                break;
            case 6:
            printf("Saturday");
                break;
        }

        printf("%d ,%d, %d, is a %d", year,month,day,total_num);
        }
        else
        {
            printf("invalid\n");
        }



    return 0;

    }

Answer 1

你说:

printf("%d ,%d, %d, is a %d", year,month,day,total_num);

这将要打印

L, M, N, is a P

其中 L、M、N 和 P 是数字。

您需要 printf() name-of-days switch 并且需要删除最后的 %d 和 printf 中的 total_num。然后 printf 将打印

L, M, N, is a

和 name-of-days switch 中的 printf 将在同一行打印出当天的名称，给你

L, M, N, is a XXXXXXXXXXX

---

编辑以解决评论:

查看您的程序的输出语句。

遇到的第一个输出语句是 name-of-days 开关中的 printf 调用，它打印出日期名称。因此，当您的程序根据您提到的输入运行时，将打印出的第一件事是

Saturday

然后在日期切换之后，下一个 printf 是

printf("%d ,%d, %d, is a %d", year,month,day,total_num);

因为year是1982，month是4，day是24，total_num为6，则printf会输出

1982, 4, 24, is a 6

和之前的Saturday输出在同一行，也就是说整个输出是

Saturday1982, 4, 24, is a 6