C不保存其他方法的更改

Question

我试图实现一个链表类在C和我有问题。我有添加新节点的=运算符。

链接列表类接口：

template <typename Type>

class LinkedList {
public:
    LinkedList<Type>* head;
//  linked list stracture
    Type data;
    LinkedList<Type>* next;
//  others ....
    size_t length;
public:
    LinkedList();
    ~LinkedList();
    void initializeHead(LinkedList<Type>* headPtr);
    size_t size() const;
    LinkedList& operator+=(const Type& add);
    void operator-=(const Type& remove);
    LinkedList<Type>& operator[] (const size_t index) const;
    bool operator== (const LinkedList<Type> &versus) const;
    friend ostream& operator<< (ostream& out,LinkedList& obj);
};

这里我有=重载实现：

template <typename Type> LinkedList<Type>& LinkedList<Type>::operator +=(const Type& add) {
    // head ptr - :)
    LinkedList<Type>* p = head->next;
    // go to the end
    while(p) p = p->next;
    // now on end - create new..!!!
    try {
        p = new LinkedList<Type>;
    } catch (bad_alloc& e) {
        cout << "There\'s an allocation error....";
    } catch (...) {
        cout << "An unknown error.." << endl;
    }// fill and done
    p->data = add;
    p->next = NULL;
    // increment length .........
    ++head->length;
    // done ............
    return *p;
}

此外，我还有“数组”访问重载方法：

template <typename Type> LinkedList<Type>& LinkedList<Type>::operator [](const size_t index) const {
    if(index < 0 || index >= length) // invaild argument
        throw  exception();
    // continue
    LinkedList<Type>* p = head;
    for(size_t i = 0; i < index; ++i) p = p->next; // we are at what we want
    return *p;
}

一切正常-我检查了dibugger，

问题是-=不在“head”中保存新节点-

有人知道为什么新的分配没有链接到头-

谢谢！！

Answer 1

正如其他答案所说，当您尝试添加时，您超出了列表。试着这样做：

template <typename Type> LinkedList<Type>& LinkedList<Type>::operator +=(const Type& add)
{
    LinkedList<Type> *last;

    // Find the last node in the list
    for (last = head; last != 0 && last->next != 0; last = last->next)
    {
    }

    // `last` now points to the last node in the list, or is zero
    // If zero (i.e. NULL) then list is empty

    if (last == 0)
    {
        head = new LinkedList<Type>;
        head->next = 0;
        head->data = add;
        head->length = 0;
    }
    else
    {
        last->next = new LinkedList<Type>;
        last->next->next = 0;
        last->next->data = add;
    }

    // We can safely use `head` as we are sure it won't be zero
    head->length++;

    // Return the added node
    return (last != 0 ? *last->next : *head);
}

Answer 2

而不是：

// go to the end
while(p) p = p->next;

您需要：

head->next = p;

Answer 3

在之后，而（p）p=p-

接下来你要做的是p=newlinkedlist