Question

这段代码有什么问题？调试后我在runner = runner->next;收到错误，但我没看到它有什么问题。

void Linkedlist<T>::add(T item)
{
    if (root == 0)
    {
        root = new Node<T>(item);
        cout << "Add, Root is empty \n";
    }
    else
    {
        cout << "Add, Root  is not empty \n";
        Node<T> * runner = root;
        while (runner != 0)
        {
            runner = runner->next;
        }
        runner = new Node<T>(item);
    }
}

Answer 1

应该是这样的：

Node<T> * runner = root;
while (runner->next != NULL)
{
    runner = runner->next;
}
runner->next = new Node<T>(item);

您的代码只更改局部变量，而不是实际节点。（请注意，runner->next是实际列表中的对象，而runner只是一个局部变量。）

Linkedlist - 模板 - 指针

1 个答案: