实验6 C语言结构体、枚举应用编程

实验任务4

代码task4.c

#include <stdio.h>
#define N 10

typedef struct {
    char isbn[20];          // isbn号
    char name[80];          // 书名
    char author[80];        // 作者
    double sales_price;     // 售价
    int  sales_count;       // 销售册数
} Book;

void output(Book x[], int n);
void sort(Book x[], int n);
double sales_amount(Book x[], int n);

int main() {
    Book x[N] = {{"978-7-229-14156-1", "源泉", "安.兰德", 84, 59},
                 {"978-7-5133-5261-1", "李白来到旧金山", "谭夏阳", 48, 16},
                 {"978-7-5617-4347-8", "陌生人日记", "周怡芳", 72.6, 27},
                 {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49},
                 {"978-7-5046-9568-0", "数据化决策", "道格拉斯·W·哈伯德", 49, 42},
                 {"978-7-5133-4388-6", "美好时代的背后", "凯瑟琳.布", 34.5, 39},
                 {"978-7-1155-0509-5", "无穷的开始:世界进步的本源", "戴维·多伊奇", 37.5, 55},
                 {"978-7-5321-5691-7", "何为良好生活", "陈嘉映", 29.5 , 31},
                 {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118, 42},
                 {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5, 44}};
    
    printf("图书销量排名: \n");
    sort(x, N);
    output(x, N);

    printf("\n图书销售总额: %.2f\n", sales_amount(x, N));
    
    return 0;
}

void output(Book x[], int n){
    int i;
    printf("isbn号\t\t\t书名\t\t\t\t作者\t\t\t售价\t销售册数\n");
    for(i = 0; i < n; i++){
        printf("%-20s\t%-30s\t%-20s\t%g\t%d\n", x[i].isbn, x[i].name, x[i].author, x[i].sales_price, x[i].sales_count);
    }
}

void sort(Book x[], int n){
    Book t;
    int i, j;
    for(i = 0; i < n-1; i++){
        for(j = 0; j < n-i-1; j++){
            if(x[j].sales_count < x[j+1].sales_count){
                t = x[j];
                x[j] = x[j+1];
                x[j+1] = t;
            }
        }
    }
}

double sales_amount(Book x[], int n){
    double avg;
    int s = 0, i;
    for(i = 0; i < n; i++){
        s += x[i].sales_price;
    }
    avg = s / n;
    return avg;
}

 

运行结果

 

实验任务5

代码task5.c

#include <stdio.h>

typedef struct {
    int year;
    int month;
    int day;
} Date;

// 函数声明
void input(Date *pd);                   // 输入日期给pd指向的Date变量
int day_of_year(Date d);                // 返回日期d是这一年的第多少天
int compare_dates(Date d1, Date d2);    // 比较两个日期: 
                                        // 如果d1在d2之前，返回-1;
                                        // 如果d1在d2之后，返回1
                                        // 如果d1和d2相同，返回0

void test1() {
    Date d;
    int i;

    printf("输入日期:(以形如2023-12-11这样的形式输入)\n");
    for(i = 0; i < 3; ++i) {
        input(&d);
        printf("%d-%02d-%02d是这一年中第%d天\n\n", d.year, d.month, d.day, day_of_year(d));
    }
}

void test2() {
    Date Alice_birth, Bob_birth;
    int i;
    int ans;

    printf("输入Alice和Bob出生日期:(以形如2023-12-11这样的形式输入)\n");
    for(i = 0; i < 3; ++i) {
        input(&Alice_birth);
        input(&Bob_birth);
        ans = compare_dates(Alice_birth, Bob_birth);
        
        if(ans == 0)
            printf("Alice和Bob一样大\n\n");
        else if(ans == -1)
            printf("Alice比Bob大\n\n");
        else
            printf("Alice比Bob小\n\n");
    }
}

int main() {
    printf("测试1: 输入日期, 打印输出这是一年中第多少天\n");
    test1();

    printf("\n测试2: 两个人年龄大小关系\n");
    test2();
}

void input(Date *pd) {
    scanf("%d-%d-%d", &(*pd).year, &(*pd).month, &(*pd).day);
}

int day_of_year(Date d) {
    int m[13] = { 0, 31,28,31,30,31,30,31,31,30,31,30,31 };
    if ((d.year % 4 == 0 && d.year % 400 == 0) || d.year % 4 == 0) {
        m[2] = 29;
    }
    int sum = 0, i;
    for ( i = 1; i <= d.month - 1; i++) {
        sum += m[i];
    }
    sum += d.day;
    return sum;
}

int compare_dates(Date d1, Date d2) {
    if(d1.year < d2.year)
        return -1;
    else if(d1.year > d2.year)
        return 1;
    else
    {
        if(day_of_year(d1) < day_of_year(d2))
            return -1;
        else if(day_of_year(d1) == day_of_year(d2))
            return 0;
        else
            return 1;
    }
}

 

运行结果

 

实验任务6

代码task6.c

 

#include <stdio.h>
#include <string.h>

enum Role {admin, student, teacher};

typedef struct {
    char username[20];  // 用户名
    char password[20];  // 密码
    enum Role type;     // 账户类型
} Account;


// 函数声明
void output(Account x[], int n);    // 输出账户数组x中n个账户信息，其中，密码用*替代显示

int main() {
    Account x[] = {{"A1001", "123456", student},
                    {"A1002", "123abcdef", student},
                    {"A1009", "xyz12121", student}, 
                    {"X1009", "9213071x", admin},
                    {"C11553", "129dfg32k", teacher},
                    {"X3005", "921kfmg917", student}};
    int n;
    n = sizeof(x)/sizeof(Account);
    output(x, n);

    return 0;
}

// 待补足的函数output()实现
// 功能：遍历输出账户数组x中n个账户信息
//      显示时，密码字段以与原密码相同字段长度的*替代显示
void output(Account x[], int n) {
    int i;
    char k[10][10] = {"admin","student","teacher"};
    for(i = 0; i < n; i++)
    {
        int j = 0;
        while(x[i].password[j] != '\0')
        {
            x[i].password[j] = '*';
            j++;
        }
        printf("%s\t\t%-9s\t\t%s\n", x[i].username, x[i].password, k[x[i].type]);    
    }
}

运行结果