Ubuntu 用C语言输入一个年份月份，判断该年该月由多少天；（要求封装+传参调用）；

以下是一个简单的C语言函数，可以用来计算给定年份月份的天数：

#include <stdio.h>

int days_in_month(int year, int month) {
    int is_leap = (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
    int days = 0;
    switch (month) {
        case 1:
            days = 31;
            break;
        case 2:
            days = is_leap ? 29 : 28;
            break;
        case 3:
            days = 31;
            break;
        case 4:
            days = 30;
            break;
        case 5:
            days = 31;
            break;
        case 6:
            days = 30;
            break;
        case 7:
            days = 31;
            break;
        case 8:
            days = 31;
            break;
        case 9:
            days = 30;
            break;
        case 10:
            days = 31;
            break;
        case 11:
            days = 30;
            break;
        case 12:
            days = 31;
            break;
        default:
            printf("Invalid month: %d\n", month);
            break;
    }
    return days;
}

int main() {
    int year, month;
    printf("Enter year: ");
    scanf("%d", &year);
    printf("Enter month: ");
    scanf("%d", &month);
    int days = days_in_month(year, month);
    printf("There are %d days in %d/%d.\n", days, year, month);
    return 0;
}

你可以在需要计算天数的地方调用 `days_in_month` 函数，并传入年份和月份作为参数。例如:

int days = days_in_month(2021, 3);
printf("There are %d days in March 2021.\n", days);

输出结果为：

There are 31 days in March 2021.