我在双向链表中遇到错误（碎片错误） [英] Im getting an error (fragmentation fault) in my doubly linked list

问题描述

void DeleteNode()
{
struct node *current;
struct node *prev;
int found = 0;
int n;
prev = allocate();
current = allocate();
current = head;
prev = NULL;

printf("\nEnter the value to eliminate: ");
scanf("%d",&n);

if(head != NULL)
{
    //Assume list: 45,56,22,65
    //We want to delete 22
    while(current != NULL && found != 1)
    {
        if(current -> data == n)
        {
            if(current == head)
                {
                    head = head -> next;
                    head -> previous = NULL;
                }
            else if(current == tail)
                {
                    prev -> next = NULL;
                    tail = prev;
                }
            else
                {
                    prev -> next = current -> next; //
                    current -> next -> previous = prev; //
                }
            printf("\n %d was succesfully deleted from the list.\n",n);
            found = 1;
        }
        prev = current; //45,56,22
        current = current -> next; //56,22,65
        
    }
    if(found == 0)
        printf("\n%d is not found in the list!\n",n);
    else
    {
        free(prev);
    }
}
else
    printf("\nThe list is empty\n");

}

我的尝试：



这是我的代码，我试图从列表中删除一个数字。这是一个双重链表。问题是当我在列表中只有1个数字并且我尝试删除它时，我得到分段错误。其余的工作正常。* /

What I have tried:

This is my code, I am trying to delete a number from the list. This is a doubly linked list. The problem is when I have only 1 number in the list and i try to delete it, i get segmentation fault. For the rest it works.*/

推荐答案

if(current == head)
    {
        head = head -> next;
        head -> previous = NULL;
    }

如果这是列表中唯一的项目，那么 head - > next 将是 NULL 。所以第二行会导致分段错误。

If this is the only item in the list then head -> next will be NULL. So the second line will cause the segmentation fault.

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