我正在实现一个类似于队列的双向链表。因此,当我将节点添加到列表(例如 5 个节点)并清空列表并尝试将新节点添加到列表时,它会给我一个分段错误(核心转储)。我不知道它为什么这样做。你可以解释吗?
链表.h
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct node{
int d;
struct node *prev;
struct node *next;
}node;
typedef struct linkedlist{
int size;
struct node *first;
struct node *last;
}linkedlist;
链表.c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "linkedlist.h"
linkedlist* createList(){
linkedlist* myList = (linkedlist*)calloc(1,sizeof(linkedlist));
myList->first = NULL;
myList->last = NULL;
myList->size =0;
return myList;
}
static node* createNode(int n){
node *myNode = (node*)calloc(1,sizeof(node));
myNode->d = n;
myNode->prev = NULL;
myNode->next = NULL;
return myNode;
}
void insertNode(linkedlist* l, int num){
node *temp, *newNode;
newNode = createNode(num);
if (l->size == 0){
newNode->next = NULL;
newNode->prev = NULL;
l->first = newNode;
l->last = newNode;
l->size++;
}
else{
temp = l->first;
while (temp->next != NULL){
temp = temp->next;
}
newNode->prev = temp;
temp->next = newNode;
newNode->next = NULL;
l->size++;
}
}
int deleteNode(linkedlist* l){
node *temp = calloc(1,sizeof(node));
if (l->first ==NULL){
return -1;
}
else if (l->size ==1){
free(l->first);
l->first= NULL;
l->last = NULL;
l->size--;
}
else if (l->size > 1){
temp = l->first;
l->first = temp->next;
free(temp);
}
}
void display(linkedlist *l){
node *temp = calloc(1,sizeof(node));
temp = l->first;
if (temp == NULL){
printf("The list is empty\n");
}
while (temp != NULL) {
printf("-> %d ", temp->d);
temp = temp->next;
}
}
int main(){
linkedlist *myList = createList();
int choice, temp=0, numb;
printf("(1) Insert \n (2) Delete \n");
for (temp; temp<10; temp++){
printf("Choice :");
scanf ("%d", &choice);
switch(choice) {
case 1: {
printf("Enter a Number: ");
scanf("%d", &numb);
insertNode(myList, numb);
display(myList);
break;
}
case 2:{
deleteNode(myList);
display(myList);
break;
}
}
}
}