[英]Linked Lists and Structures in C - Passing Linked List and Structures as Parameters in function (C)
我有以下代碼,並且我試圖使其更加動態和可重復使用。 好吧,我有一個名為Student和struct list的結構,其中包含所有添加的學生。 我有一個函數“ int addStudent(Student b,list StudentList){”,並且我試圖將結構Student和StudentList作為參數傳遞。 但是問題是我做錯了事,我的名單沒有包含所有添加的學生。 它僅包含最后一個。 你能幫助我嗎?
注意:我必須為"int addStudent(Student b, list StudentList)"
創建一個主體。 不允許更改此函數的聲明 ...這對我來說非常困難,我需要一些建議來進行工作...
先感謝您!
#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;
}
addStudent方法始終覆蓋先前的節點。 因此,您的列表僅包含1個節點。 另外,要“存儲”鏈接列表,您需要保持指向列表頭(第一個元素)的指針。
花一會兒用指針轉到最后一個元素,然后在列表中創建新元素void addStudent(Student *b, list *StudentList) list *elem; elem = StudentList; while (elem->next) elem = elem->next; //now you can create your elem elem->next = malloc(sizeof(list)); elem->next->student = b; elem->next->next = NULL;
void addStudent(Student *b, list *StudentList) list *elem; elem = StudentList; while (elem->next) elem = elem->next; //now you can create your elem elem->next = malloc(sizeof(list)); elem->next->student = b; elem->next->next = NULL;
您有兩個問題:
1)您將局部變量的地址添加到列表中
2)您永遠不會展開列表,即它始終只包含一個元素
解決問題1)更改addStudent
函數,如下所示:
int addStudent(Student* b, list* StudentList){
^^^ ^^^
Use a pointer
StudentList->Student = b;
^^^
Save the pointer
return 1;
}
並這樣稱呼:
int x=addStudent(&b, StudentList);
^^
解決問題2):
您需要malloc
一個新的列表項並將其插入到當前列表中。
但是,您當前的代碼有點奇怪,因為您還在main
分配了一個list
,但沒有放入任何內容。 相反,似乎最好只在addStudent
函數中分配list
項。
一個簡單的方法是:
// This function inserts new item in the front of the list
list* addStudent(Student* b, list* StudentList){
list* t;
t = malloc(sizeof(list));
if (!t) exit(1);
t->next = StudentList;
t->Student = b;
return t;
}
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 = addStudent(&b, StudentList);
StudentList = addStudent(&c, StudentList);
printlist(StudentList);
return 0;
}
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