实验6_C语言结构体、枚举应用编程

#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;
}

// 待补足：函数output()实现
void output(Book x[], int n){
    int i;
    printf("%-20s%-30s%-20s%-10s%s\n", "ISBN号", "书名", "作者", "售价", "销售册数"); 
    for(i = 0; i < n; i++){
        printf("%-20s%-30s%-20s%-10g%d\n", x[i].isbn, x[i].name, x[i].author, x[i].sales_price, x[i].sales_count); 
    }
    
}


// 待补足：函数sort()实现
void sort(Book x[], int n){
    Book y[1];
    int i, j;
    for(i = 0; i < n; i++){
        for(j = 0; j < n-1-i; j++){
            if(x[j].sales_count < x[j+1].sales_count){
                y[0] = x[j];
                x[j] =  x[j+1];
                x[j+1] = y[0]; 
            }
        }
    }
}


// 待补足：函数sales_count()实现
double sales_amount(Book x[], int n){
    double sum = 0.0;
    int i;
    for(i = 0; i < n; i++){
        sum = sum + x[i].sales_price * x[i].sales_count;
    }
    return sum;
}

#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, j, a;
    char p[10][20];
    for(i = 0; i < n; i++){
        a = strlen(x[i].password);
        printf("%-20s", x[i].username);
            for(j = 0; j < a; j++){
                p[i][j] = '*';
            }
        p[i][j] = '\0';
        printf("%-20s", p[i]);
        if (x[i].type == 0){
            printf("%s","admin\n");
        }
        else if (x[i].type == 1){
            printf("%s","student\n");
        }
        if (x[i].type == 2){
            printf("%-s","teacher\n");
        }
    }
}