使用析构函数释放链接对象 [英] Using the destructor to free linked objects

问题描述

我有一个名为node的类。我将一堆节点对象链接在一起以形成链表。当节点析构函数被调用时，它只删除第一个节点。如何迭代整个链接的节点列表并删除每个节点对象？

I have a class called "node". I link a bunch of node objects together to form a linked list. When the "node" destructor is called, it only deletes the first node. How do I iterate through the entire linked list of nodes and delete each node object?

这里是类定义：

class Node
{
private:
double coeff;
int exponent;
Node *next;
public:

Node(double c, int e, Node *nodeobjectPtr)
{
    coeff = c;
    exponent = e;
    next = nodeobjectPtr;
}

~Node()
{
    printf("Node Destroyed");
}

通过在指向第一个节点的指针上调用delete来调用析构函数链接节点列表。

The destructor is called by invoking delete on the pointer to the first node of the linked node list.

推荐答案

如何遍历整个节点链表，节点对象？

How do I iterate through the entire linked list of nodes and delete each node object?

这将是 cleaner ，如果你有一个单独的类来管理整个列表，这样节点可以是简单的数据结构。然后你只需要一个简单的循环列表的析构函数：

It would be cleaner if you had a separate class to manage the entire list, so that nodes can be simple data structures. Then you just need a simple loop in the list's destructor:

while (head) {
    Node * victim = head;
    head = victim->next;    // Careful: read this before deleting
    delete victim;
}

如果你真的想将列表管理委托给节点本身，需要更加小心：

If you really want to delegate list management to the nodes themselves, you'll need to be a bit more careful:

while (next) {
    Node * victim = next;
    next = victim->next;
    victim->next = nullptr;  // Careful: avoid recursion
    delete victim;
}

在此方案下，您还需要在删除节点时从列表中删除它后 - 再次，请确保您重置其指针，以便它不会删除列表的其余部分。这是另一个支持单独的列表类的原因。

Under this scheme, you'll also need to be careful when deleting a node after removing it from the list - again, make sure you reset its pointer so it doesn't delete the rest of the list. That's another reason to favour a separate "list" class.

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