如何解决从双向链表中删除节点会产生分段错误-C
我编写了一个程序，可以将char字符添加到双向链接列表的开头。现在，有了这个列表，我的程序的目的就是从列表中完全删除某些字符。例如（使用大括号仅用于代表目的）：如果列表由 {a，b，b，a，c} 组成，则我的程序可以从列表中删除所有“ b”以使其成为 {a，a，c} 。此外，如果我的列表是 {b，a，c，a} 或 {a，c，a，b} ，并且我要删除“ b”，则程序在两种情况下都可以正常运行，并给我 {a，c，a} 。
但是有很多问题（在所有情况下，我都想删除“ b”）：
- 如果我的列表是 {b，a，b，a，c} （前面的“ b”和中间的某处），则会出现细分错误（我认为这与使用游标在while循环中，但我不知道为什么以及如何解决它）
- 如果我的列表是 {a，b，b，a，c，b} （中间是“ b”，最后是“ b”），那么输出将给出奇怪的符号（我假设它的a内存故障，不知道为什么）
这是我正在使用的代码：
#include<stdio.h>
#include<stdlib.h>
struct list
{
int data;
struct list* next;
struct list* prev;
};
struct list* head; // global variable - pointer to head node of list.
struct list* last; // global variable - pointer to last node of list
//Creates a new list and returns pointer to it.
struct list* GetNewNode(char x)
{
struct list* newNode
= malloc(sizeof(struct list));
newNode->data = x;
newNode->prev = NULL;
newNode->next = NULL;
return newNode;
}
//Inserts a list at head of doubly linked list
void InsertAtHead(char x)
{
struct list* newNode = GetNewNode(x);
if(head == NULL)
{
head = newNode;
return;
}
head->prev = newNode;
newNode->next = head;
head = newNode;
struct list* temp = head;
while (temp->next != NULL) temp = temp->next;
last = temp;
}
void remove_element (char character)
{
struct list * cursor,*previous,*store_el;
//int boolean = 0;
if (head == NULL) return;
else
{
cursor = head;
while(cursor != NULL)
{
if (cursor->data == character)
{
if (cursor->prev == NULL)
{
// printf("deleting from front\n");
previous = head;
head = head->next;
head->prev = NULL;
//boolean = 1;
//free(previous);
}
if (cursor->next == NULL)
{
//printf("deleting from back\n");
previous = last;
last = last->prev;
last->next = NULL;
//boolean = 1;
//free(previous);
}
else
{
// printf("deleting from middle\n");
previous = cursor;
cursor = cursor->next;
cursor->prev = previous->prev;
store_el = previous->prev;
store_el->next = cursor;
cursor = head;
}
free(previous);
//printf("head data = %c\n",cursor->data);
}
cursor = cursor->next;
}
}
}
//Prints all the elements in linked list in forward traversal order
void Print()
{
struct list* temp = head;
printf("Forward: ");
while(temp != NULL)
{
printf("%c ",temp->data);
temp = temp->next;
}
printf("\n");
}
int main()
{
char character;
/*Driver code to test the implementation*/
head = NULL; // empty list. set head as NULL.
// Calling an Insert and printing list before and after deletion of character
InsertAtHead('c');
InsertAtHead('a');
InsertAtHead('b');
InsertAtHead('b');
InsertAtHead('a');
Print();
printf("After deletion:\n");
remove_element ('b');
Print();
}
我花了7到8个小时来尝试完成这段代码（大约3个小时来查找代码问题）。附带说明：如果有任何技巧可以提高编码效率，我将不胜感激，例如，有什么可以帮助您提高效率的技巧等。
解决方法
/*
*I changed the name of your variable 'last' to 'tail'
*I removed the code at the end of your InsertAtHead function
*I added "tail = newNode;"
*I changed the name of your variable 'previous' to 'garbage'
*I removed your variable 'store_el' completely.
*I could have changed the whole code in your remove element function because the 3 cases are unnecessary but anyway.
*/
//Inserts a list at head of doubly linked list
void InsertAtHead(char x){
struct list* newNode = GetNewNode(x);
if (head == NULL){
head = newNode;
tail = newNode;
return;
}
head->prev = newNode;
newNode->next = head;
head = newNode;
}
void remove_element (char character){
struct list * cursor,*garbage;
cursor = head;
while(cursor != NULL){
if (cursor->data == character){
garbage = cursor;
if (cursor->prev == NULL){
head = head->next;
If (head!=NULL) head->prev = NULL;
cursor=head;
}else if (cursor->next == NULL){
tail = tail->prev;
tail->next = NULL;
cursor=NULL;
}else{
garbage->prev->next = garbage->next;
garbage->next->prev = garbage->prev;
cursor=cursor->next;
}
free(garbage);
} else cursor=cursor->next;
}
}
立即尝试。
代码的问题是您正在使用释放的内存。
/*
*cursor and previous point to the same memory address
*you free the memory that the variable previous points so the cursor points to that freed memory
*when you save the next address to the cursor using that freed memory you create an undefined behaviour (your code may work or may not)
*/
cursor = head;
while(cursor != NULL)
{
if (cursor->data == character)
{
if (cursor->prev == NULL)
{
previous = head;
head = head->next;
head->prev = NULL;
free(previous);
...
cursor=cursor->next;
改进代码：
#include<stdio.h>
#include<stdlib.h>
//The type of your variable data was wrong. I changed it to char
struct list{
char data;
struct list *prev,*next
};
void remove_element (char character){
struct list * cursor,*garbage;
cursor = head;
while(cursor != NULL){
if (cursor->data == character){
garbage = cursor;
if (garbage->prev!=NULL ) garbage->prev->next = garbage->next;
if (garbage->next!=NULL ) garbage->next->prev = garbage->prev;
cursor=cursor->next;
if (head==garbage) head=cursor;
//Basically the tail variable has no use for your current program.
//if (tail==garbage) tail=garbage->prev;
free(garbage);
} else cursor=cursor->next;
}
}
尝试更新光标
cursor = head;
从正面删除后。
,Ellothere我认为您的程序中的问题是，您以前在分配节点的地址，但没有并排释放它。
if (cursor->prev == NULL)
{
// printf("deleting from front\n");
previous = head;
head = head->next;
head->prev = NULL;
//boolean = 1;
//free(previous);
}
if (cursor->next == NULL)
{
//printf("deleting from back\n");
previous = last;
last = last->prev;
last->next = NULL;
//boolean = 1;
//free(previous);
}
这里您要存储节点，但最后要释放它，然后有2个if语句，那么正在发生的事是{abbac}中的'b'彼此相邻，因此首先 previous 变量存储的是第一个'b'的地址，然后存储的是 next 'b'的地址，它只是释放了'b'和那个'b'仍然在那里。简而言之，您应该并排释放节点。我做了这个小的更改，效果很好。
if (cursor->prev == NULL)
{
// printf("deleting from front\n");
previous = head;
head = head->next;
head->prev = NULL;
free(previous);
//boolean = 1;
//free(previous);
}
if (cursor->next == NULL)
{
//printf("deleting from back\n");
previous = last;
last = last->prev;
last->next = NULL;
free(previous);
//boolean = 1;
//free(previous);
}
else
{
// printf("deleting from middle\n");
previous = cursor;
cursor = cursor->next;
cursor->prev = previous->prev;
store_el = previous->prev;
store_el->next = cursor;
cursor = head;
}
我刚刚在两个 if 语句中添加了 免费 函数。
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