C（链表）中的结构内部结构

Question

我正在尝试编写一个程序，该程序允许我按顺序打印教师的信息，包括学生 ID。 例如：当我尝试运行程序时，output 将类似于以下内容：

Teacher Apple, Age 55:
Student ID: 77
Student ID: 5456
Student ID: 4729
Student ID: 3784

Teacher bob, Age 67:
Student ID: 477
Student ID: 546
Student ID: 429
Student ID: 784

.. 等等。

但是，我想不出一种以正确格式打印它们的方法。 我在使用嵌套循环时遇到问题。 这是我的代码：

struct Teacher {

    char* name;
    int age;
    struct Student* stds;
    struct Teacher* next;
};

struct Student 
{
    int id;
    struct Student* next;
};


int main() {

    //aray to store teachers' names
    char names[3][10] = { "first", "second", "third" };
//their age
    int age[3] = { 1,2,3 };
//students' id
    int id[4] = { 1, 2, 3, 4 };
//alocate the linkedlist in mem
    struct Teacher* head = (struct Teacher*)malloc(sizeof(struct Teacher));
    struct Student* headstd = (struct Student*)malloc(sizeof(struct Student));
    
    
    head->next = NULL;
    headstd->next = NULL;
    //head ->next = (struct Teacher*)malloc(sizeof(struct Teacher));
//adding temporary head and modefied it so I don't touch the actual head 
    struct Teacher* temhead = (struct Teacher*)malloc(sizeof(struct Teacher));
    struct Student* stdhead = (struct Student*)malloc(sizeof(struct Student));
    //head->name = 
    //(char)first.name ="first";
    temhead->next = NULL;
    stdhead->next = NULL;
///loop thorough the teachers
    for (int i = 0; i < 3; i++) {
//loop through the students 
        for (int j = 0; j < 4; j++) {
            stdhead->next = (struct Student*)malloc(sizeof(struct Student));
            stdhead = stdhead->next;
            stdhead->id = *(id + j);
            //temhead->age = 44;
            
            printf("Student ID: %d \n ", stdhead->id);
        }
//here after it goes through the students, it goes through teachers 
        temhead->next = (struct Teacher*)malloc(sizeof(struct Teacher));
        temhead = temhead->next;
//here I assigned the name of teachers to the array I have implementd @ first (names[3][10]
        temhead->name = *(names + i);
//print names (for now "first, second, ..")
        printf("node: %s \n ", temhead->name);

    }
//
//  for (int i = 0; i < 3; i++) {
//      temhead->next = (struct Teacher*)malloc(sizeof(struct Teacher));
//      //temhead->stds = (struct Teacher*)malloc(sizeof(struct Teacher));
//      temhead = temhead->next;
//      temhead->age = *(age + i);
//      //temhead->age = 44;
//      printf("%d \n ", temhead->age);
//
//  }
//
//  
//}

我刚开始学习 C，我真的很感激任何帮助。 我的输出如下所示：

Answer 1

打印代码可能看起来像这样。 但是，除非您首先正确地组装教师和学生，否则它不会起作用。

for (struct Teacher *t = head; t != NULL; t = t->next)
{
    printf("Teacher %s, Age %d:\n", t->name, t->age);
    for (struct Student *s = t->stds; s != NULL; s = s->next)
        printf("Student ID: %d\n", s->id);
}

为了组合这些列表，您可能需要编写一些函数来创建教师并将其添加到列表中，并将学生添加到教师中。 然后首先调用这些函数来组装列表。