通过一定的准则删除链表元素 [英] Delete element of linked list by a certain criterion

问题描述

我已经写了这个功能：

List* delPaintingCode(List* head, char *code)
{
    List *p,*q;

    for(p=head;p!=NULL;q=p,p=p->next)
    {
        if(!strcmp(code,p->code))
        {
            if (p==head)
            {
                q=head;
                head=p->next;
                free(q);
            }
            else
            {
                q->next=p->next;
                free(p);
                p=q;
            }
        }
    }
    return head;
}

，当我把它在另一个功能：

and when I call it in another function:

void delpainting()
{ 
    char code[50];
    printf("code ");
    scanf("%s",code);

    List *head=NULL;

    head=delPaintingCode(filetolist,code);
}

程序崩溃等等。我有一个警告：通过delPainting code的参数1从兼容的指针类型

所以，我应该怎么字符串传递给函数？

So how should I pass the string to the function?

推荐答案

这导致崩溃的逻辑问题是的如果（P ==头）枝您code：当你删除了最初的元素，释放头而不更新 p 。这导致在很下一次迭代解引用一个释放的指针。

The logical problem that leads to a crash is in the if (p==head) branch of your code: when you delete the initial element, you free the head without updating p. This leads to dereferencing a freed pointer on the very next iteration.

您可以通过引入一个假的节点与头在接下来，并返回<$解决问题C $ C>接下来，就像这样：

You can fix the problem by introducing a fake node with head in its next, and returning the next, like this:

List fake;
fake.next = head;
// This loop always looks ahead by one element, i.e. at p->next.
for(List *p = &fake ; p->next != NULL ; p = p->next) {
    if(strcmp(code, p->next->code)) {
        continue;
    }
    List *q = p->next;
    p->next = q->next;
    free(q);
}
return fake.next;

本方法适用于初始元素，也因为我们增加了一个假头我们的名单，所以第一次围绕 P-＆gt;接下来是一样的头。这让我们统一治疗头元素和所有其他元素。

This approach works for the initial element, too, because we added a fake head to our list, so the first time around p->next is the same as head. This lets us unify treatment of head element and all other elements.

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