#include <stdio.h>
#include <stdlib.h>
typedef struct student {
int id;
char name[200];
char gender;
int age;
struct student *next;
} Student;
int main(void) {
Student *head = NULL, *tail = NULL;
// 从键盘输入4个学生的数据
for (int i = 0; i < 4; i++) {
// 动态分配内存
Student *node = (Student*)malloc(sizeof(Student));
if (node == NULL) {
printf("Memory allocation failed.\n");
exit(1);
}
// 从键盘输入学生数据
printf("Enter student %d data:\n", i+1);
scanf("%d%s %c%d", &node->id, node->name, &node->gender, &node->age);
// 将新节点插入链表
if (head == NULL) {
// 如果链表为空，则新节点为链表的第一个节点
head = node;
tail = node;
} else {
// 将新节点插入到链表的最后一个节点之后
tail->next = node;
tail = node;
}
}
// 从键盘输入一个年龄
int age;
printf("Enter an age: ");
scanf("%d", &age);
// 删除链表中所有包含此年龄的节点
Student *curr = head, *prev = NULL;
while (curr != NULL) {
if (curr->age == age) {
// 如果找到了包含此年龄的节点，则从链表中删除它
if (prev == NULL) {
// 如果当前节点是链表的第一个节点，则将头指针指向下一个节点
head = curr->next;
} else {
// 将前一个节点的 next 指针指向当前节点的下一个节点
prev->next = curr->next;
}
// 释放当前节点的内存
free(curr);
// 将当前节点更新为前一个节点
curr = prev;
}
// 更新 prev 和 curr 指针
prev = curr;
curr = curr->next;
}
// 输出剩余节点的数据
for (curr = head; curr != NULL; curr = curr->next) {
printf("%d %s %c %d\n", curr->id, curr->name, curr->gender, curr->age);
}
return 0;
}
这个代码还不能执行成功。我想知道是看看哪里有问题。
参考代码实现如下，望采纳
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_NAME_LEN 20
typedef struct student {
int id;
char name[MAX_NAME_LEN+1];
char gender;
int age;
struct student *next;
} Student;
Student *create_student(int id, const char *name, char gender, int age) {
Student *s = (Student *)malloc(sizeof(Student));
if (s == NULL) {
printf("Error: failed to allocate memory for student.\n");
exit(1);
}
s->id = id;
strncpy(s->name, name, MAX_NAME_LEN);
s->name[MAX_NAME_LEN] = '\0';
s->gender = gender;
s->age = age;
s->next = NULL;
return s;
}
void insert_student(Student **head, Student *s) {
if (*head == NULL) {
*head = s;
} else {
Student *p = *head;
while (p->next != NULL) {
p = p->next;
}
p->next = s;
}
}
void delete_student(Student **head, int age) {
Student *p = *head;
Student *prev = NULL;
while (p != NULL) {
if (p->age == age) {
if (prev == NULL) {
*head = p->next;
} else {
prev->next = p->next;
}
free(p);
p = prev;
}
prev = p;
p = p->next;
}
}
void print_students(const Student *head) {
const Student *p = head;
while (p != NULL) {
printf("%d %s %c %d\n", p->id, p->name, p->gender, p->age);
p = p->next;
}
}
int main() {
Student *head = NULL;
int id;
char name[MAX_NAME_LEN+1];
char gender;
int age;
for (int i = 0; i < 4; i++) {
scanf("%d", &id);
scanf("%s", name);
scanf(" %c", &gender);
scanf("%d", &age);
Student *s = create_student(id, name, gender, age);
insert_student(&head, s);
}
scanf("%d", &age);
delete_student(&head, age);
print_students(head);
return 0;
}
修改完善如下，供参考：
#include <stdio.h>
#include <stdlib.h>
typedef struct student {
int id;
char name[200];
char gender;
int age;
struct student *next;
} Student;
int main(void)
{
Student *head = NULL, *tail = NULL;
// 从键盘输入4个学生的数据
for (int i = 0; i < 4; i++) {
// 动态分配内存
Student *node = (Student*)malloc(sizeof(Student));
if (node == NULL) {
printf("Memory allocation failed.\n");
exit(1);
}
// 从键盘输入学生数据
printf("Enter student %d data:\n", i+1);
scanf("%d %s %c %d", &node->id, node->name, &node->gender, &node->age);
//scanf("%d%s %c%d", &node->id, node->name, &node->gender, &node->age);
// 将新节点插入链表
if (head == NULL) {
// 如果链表为空，则新节点为链表的第一个节点
head = node;
tail = node;
} else {
// 将新节点插入到链表的最后一个节点之后
tail->next = node;
tail = node;
}
}
// 从键盘输入一个年龄
int age;
printf("Enter an age: ");
scanf("%d", &age);
// 删除链表中所有包含此年龄的节点
Student *curr = head, *prev = NULL;
while (curr != NULL) {
if (curr->age == age)
{
// 如果找到了包含此年龄的节点，则从链表中删除它
if (prev == NULL) {
// 如果当前节点是链表的第一个节点，则将头指针指向下一个节点
head = curr->next;
free(curr);
curr = head;
} else {
// 将前一个节点的 next 指针指向当前节点的下一个节点
prev->next = curr->next;
//}
// 释放当前节点的内存
free(curr);
// 将当前节点更新为前一个节点
curr = prev;
}
}
else{ // 更新 prev 和 curr 指针
prev = curr;
curr = curr->next;
}
}
// 输出剩余节点的数据
for (curr = head; curr != NULL; curr = curr->next) {
printf("%d %s %c %d\n", curr->id, curr->name, curr->gender, curr->age);
}
return 0;
}