我有以下代码,我正在努力使其更具动态性和可重用性。 好吧,我有一个名为 Student 的结构和一个结构列表,其中包含所有添加的学生。我有一个函数“int addStudent(Student b, list StudentList){”,我试图将结构 Student 和 StudentList 作为参数传递。但问题是我做错了什么,我的列表不包含所有添加的学生。它只包含最后一个。 你能帮帮我吗?
注意:我必须为 "int addStudent(Student b, list StudentList)"
创建主体。不允许更改此函数的声明 ...这对我来说非常困难,我需要建议来处理...
提前致谢!
#include <stdio.h>
#include <stdlib.h>
#define MAXSTRING 100
#define MAXLessonS 100
typedef enum genders{
female,
male
} genders;
typedef struct Student
{
char name[MAXSTRING];
char Surname[MAXSTRING];
enum genders gender;
int id;
char Lessons[MAXLessonS][MAXSTRING];
} Student;
typedef struct list
{
struct list * next;
struct Student * Student;
} list;
void printlist(list * StudentList)
{
list * current = StudentList;
while (current != NULL) {
printf("Student ID = %d\n", current->Student->id);
printf("Student name = %s\n", current->Student->name);
printf("Student Surname = %s\n", current->Student->Surname);
printf("Student gender = %d\n", current->Student->gender);
printf("Student Lesson = %s\n", current->Student->Lessons);
current = current->next;
}
}
int main()
{
Student b={"name 1","Surname 1",male,22,{"Lesson 1"}};
Student c={"name 2","Surname 2",female,32,{"Lesson 2"}};
list* StudentList = NULL;
StudentList = malloc(sizeof(list));
StudentList->next = NULL;
//StudentList->next->next = NULL;
int x=addStudent(b,StudentList);
StudentList->next=NULL;
int xx=addStudent(c,StudentList);
printlist(StudentList);
return 0;
}
int addStudent(Student b, list StudentList){
//StudentList=malloc(sizeof(list));
StudentList.Student = &b;
//StudentList.next->next=NULL;
//free(StudentList);
return 1;
}
最佳答案
addStudent 方法总是覆盖以前的节点。所以你的列表只包含 1 个节点。此外,要“存储”链接列表,您可能希望保留指向列表头部(第一个元素)的指针。
关于c - C 中的链表和结构 - 在函数中将链表和结构作为参数传递(C),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44286735/