我在写一个链表,我想要一个结构的析构函数(一个Node结构体)来简单地删除自己,没有任何副作用。我想要我的列表的析构函数在自身上迭代地调用Node析构函数(暂时存储下一个节点),如下所示:
我的列表类有第一个和最后一个指针 //和我的节点都有一个指向前一个和下一个 //节点 DoublyLinkedList ::〜DoublyLinkedList { Node * temp = first(); while(temp-> next()!= NULL) { delete temp; temp = temp-> next(); } }所以这将是我的Node析构函数:
Node ::〜Node { delete this;这是可以接受的,特别是在这种情况下吗? 解决方案如果正在调用Node析构函数,那么它已经在被删除的过程中。所以删除在你的Node析构函数中没有意义。
这也是错误的:
while(temp-> next()!= NULL) { delete temp; temp = temp-> next(); }而是应该将temp-> next否则,您正在访问已删除的内存。
更像这样:
DoublyLinkedList ::〜DoublyLinkedList { Node * temp = first(); while(temp!= NULL) { Node * temp2 = temp-> next(); delete temp; temp = temp2; } }
I'm writing a linked list and I want a struct's destructor (a Node struct) to simply delete itself, and not have any side effects. I want my list's destructor to iteratively call the Node destructor on itself (storing the next node temporarily), like this:
//my list class has first and last pointers //and my nodes each have a pointer to the previous and next //node DoublyLinkedList::~DoublyLinkedList { Node *temp = first(); while (temp->next() != NULL) { delete temp; temp = temp->next(); } }So this would be my Node destructor:
Node::~Node { delete this; }Is this acceptable, especially in this context?
解决方案If the Node destructor is being called, then it's already in the process of being deleted. So a delete doesn't make sense inside your Node destructor.
Also this is wrong:
while (temp->next() != NULL) { delete temp; temp = temp->next(); }Instead you should get temp->next() into a temp variable. Otherwise you are accessing deleted memory.
So more like this:
DoublyLinkedList::~DoublyLinkedList { Node *temp = first(); while (temp != NULL) { Node *temp2 = temp->next(); delete temp; temp = temp2; } }
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是否可以使用“删除此”删除当前对象?
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