如何在C ++中正确删除链接列表的节点 [英] How do I properly delete nodes of linked list in C++

问题描述

我感觉好像我实际上并没有删除该节点并释放内存.我想我只是在移动指针，所以当我打印链接列表时，列表不会打印我删除的元素.所以我的问题是我实际上是删除节点还是只是简单地重新排列指针，以使我看起来像是在删除节点(本质上只是断开链接，而不是删除节点)?谢谢您的帮助.

I feel as if I am not actually deleting the node and freeing up memory. I think I am just moving pointers around so when I print the linked list the list doesn't print the element I deleted. So my question is am I actually deleting the node or am I just simply rearranging the pointers so it looks like I am deleting the nodes(Essentially just breaking the links but not deleting the node)? Thank you for any help.

void SLL::deleteNode(int target){
Node *current = new Node;
Node *previous = new Node;

for (current = front->next, previous = front; current != NULL; current = current->next, previous=previous->next){
    if (previous->data == target && previous == front){
        front = previous->next;
        delete[] previous;
        return;
    //This if statement deletes the element if its the front
    }

    else {

        if (previous->data == target && previous->next == NULL){
            previous = NULL;
            delete[] current;
            return;
        //This if statement deletes the node if it is the back
        }


        else if (current->data==target)
        {
            previous->next = current->next;
            delete[] current;
            return;
        //This if statement deletes a node if it is in the middle
        }
    }
    }

    delete[] current;
    delete[] previous;
}

推荐答案

Node *current  = new Node;
Node *previous = new Node;

此代码导致内存泄漏-您永远不会删除此内存.您可以在不分配内存的情况下声明指针:

This code causes memory leaks - you are never deleting this memory. You can declare pointers without memory allocation:

Node *current  = nullptr;
Node *previous = nullptr;

delete将删除指针的内存，因此您实际上将删除节点. 但是对Node*使用 delete [] 不正确，它只能用于数组-用new[]分配的内存.使用不当会导致不确定的行为. 因此，要正确删除节点，请使用运算符 delete 将其删除.

delete will delete the memory of the pointer so you will actually delete Nodes. But using delete[] for the Node* is incorrect, it should be used only for arrays - the memory allocated with new[]. Improper use leads to undefined behaviour. So, to properly delete nodes delete them with operator delete.

使用内存泄漏检测工具来了解程序中是否存在内存泄漏.

Use memory leaks detection tools to know are there memory leaks in you program.

删除列表元素的代码:例如，我们有pHead指向列表的开头 (但是如果您自己编写这样的东西，它将给您带来更多的好处):

The code to delete a list element: say, we have pHead which points to the head of the list (but it would give you much more if you write such things yourself):

Node* pCur  = pHead;
Node* pPrev = pCur;

while (pCur && pCur->data != target) {
    pPrev = pCur;
    pCur  = pCur->next;
}

if (pCur==nullptr)  // not found
   return NOT_FOUND;

if (pCur == pHead) {   // first element matches
    pHead = pCur->next;
} else {
    pPrev->next = pCur->next;
}

// pCur now is excluded from the list

delete pCur;     // deallocate its memory

---

替代使用指针到指针(社区添加)

---

Alternative Using Pointer To Pointer (Community Addition)

当您使用列表中的实际指针 进行枚举时，上述内容可能会焕然一新.下面的操作从为pp分配了头指针的地址(不是它指向的节点；实际的指针本身)开始.我们遍历列表，直到pp保留指向要删除目标的节点的指针的地址(可以是头部指针，可以是某个节点中的next指针，没有区别).将要寻址的指针设置为其自己节点的next值，然后删除目标节点.

The above can take on new light when you use the actual pointers in the list to perform the enumeration. The following starts with pp being assigned the address of the head pointer (not the node it points to; the actual pointer itself). We walk the list until pp hold the address of a pointer that is pointing to a node with the target to delete (could be the head pointer, could be a next pointer in some node, makes no difference). The pointer being addressed is set to its own node's next value, then the target node is removed.

这确实应该在调试器中进行观察，以了解其工作原理，但是鉴于实际发生的情况，该算法非常简单:

This really should be watched in a debugger to see how it works, but the algorithm is remarkably simple given what is really going on:

Node **pp = &pHead;
while (*pp && (*pp)->data != target)
    pp = &(*pp)->next;

if (*pp)
{
    Node *victim = *pp;
    *pp = victim->next;
    delete victim;
}

多数民众赞成在其中.这样一来，您就可以去除头节点，而无需特殊情况下免费使用.希望这也会有所帮助.

Thats all of it. And you get head-node removal without having to special case it for free. Hope this helps as well.

这篇关于如何在C ++中正确删除链接列表的节点的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！