C语言作业（三）

一点小记录

大一上学期学到指针与结构体，下个学期就是痛苦的数据结构
指针的最基本运用还是比较简单的无非就是*&
结构体也比较易懂，联合指针与数组在程序中存读据方便了很多
联和前面所学的东西对于初学者最大的难点貌似还是理解问题，暂时还不涉及到复杂的算法

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指针输出结构体

利用指针输出一年月份的天数

#include<stdio.h>
struct mon_day
{
    char month[10];
    int day;
};

struct mon_day mon[12] = {{"January",31},{"February",28},{"March",31},{"April",30},
{"May",31},{"June",30,},{"July",31},{"August",31},{"September",30},
{"October",31},{"November",30},{"December",31}};

int main(void)
{
    struct mon_day *p;

    for ( p = mon; p < mon+12; p++)
    {
        printf("%-10s have %-2d days.\n",p -> month,p -> day);
        //结构体数组在内存地址的排列是按顺序的，只需要调用指针指向下一个内存地址即可
        //要调用结构体内的数据必须要指向运算符"->"
    }
}

---

输出天数

输入一个日期，并且输出这天是今年的多少天

#include<stdio.h>
struct date
{
    int year;
    int month;
    int day;
};

int main()
{
    struct date today;
    struct date *p = &today;

    printf("Please Enter date:");
    scanf("%d/%d/%d",&p -> year,&p -> month,&p -> day);

    printf("Today is the %d day",when(today));
}

int when (struct date d)
{
    int month[12] = {31,28,31,30,31,30,31,31,30,31,30,31};//每个月份的天数
    int days = 0;

    if(leap(d.year)) month[1] = 29;//调用自定义函数leap传入d.year数值判断是否闰年

    for (size_t i = 0; i < d.month - 1; i++)
    {
        days += month[i];
    }
  
    days += d.day;
    return days;
}

int leap(int year)//判断闰年
{
    if(year % 4 == 0 && year % 100 != 0 || year % 400 == 0) return 1;
    else return 0;
}