Question

这是我正在处理的LinkedList实现。它适用于任何数据类型，但当我尝试创建一个具有链表的类型的链接列表时出现问题我用visual studio调试指定给RtlValidateHeap的无效地址（00790000,007B16D0）

代码如下：

typedef unsigned long int LENGTH_T;
template < typename type >
struct nodes
{
    type _value;
    nodes<type> * _next_node;
    nodes() { _next_node = nullptr; }
};

template < typename type >
class LinkedList
{
    nodes<type> * _elem_nodes;
    LENGTH_T _size;
public:
    nodes<type> * node_at(LENGTH_T at);
    type& operator[] (LENGTH_T at);
    void push_back(const type src);
    LENGTH_T size() const { return _size; }
    LinkedList();
    ~LinkedList();
};

template<typename type>
nodes<type>* LinkedList<type>::node_at(LENGTH_T at) {
    if (at == 0)
        return _elem_nodes;
    else if (at > _size - 1 || _size == 0) {
        PRINT_ERROR("try to access out of range");
    }

    // tmp node for storing sequential nodes
    nodes<type> * cur_tmp_node_ptr = _elem_nodes->_next_node;

    for (size_t i = 1; i < at; i++)
        cur_tmp_node_ptr = cur_tmp_node_ptr->_next_node;

    return cur_tmp_node_ptr;
}

template<typename type>
type & LinkedList<type>::operator[](LENGTH_T at)
{
    return node_at(at)->_value;
}

template<typename type>
void LinkedList<type>::push_back(const type src)
{
    if (_size == 0) {
        _elem_nodes->_value = src;
        _size++;
    }
    else {
        nodes<type> * new_node = new nodes<type> ;
        new_node->_value = src;
        new_node->_next_node = nullptr;
        node_at(_size - 1)->_next_node = new_node;
        _size++;
    }
}

template<typename type>
LinkedList<type>::LinkedList()
{
    _size = 0;
    _elem_nodes = new  nodes<type>;
    _elem_nodes->_value = type();
    _elem_nodes->_next_node = nullptr;
}

template<typename type>
LinkedList<type>::~LinkedList()
{
    if (_size > 1) // When size = 0 , so _size-1 = -1 but _size is unsigned;
        for (LENGTH_T i = _size - 1; i > 0; i--) {
            delete ( node_at(i) );
        }
    delete ( _elem_nodes );
}

这是一个可以观察到指定问题的代码示例 f.e

struct test {
    int anything;
};

struct test2 {
    LinkedList<test> t;
};

int main()
{
    LinkedList<test2> t;
    t.push_back(test2());
    t.push_back(test2());
    return 0;
}

**编辑：我编写了自定义赋值运算符和复制构造函数，但不再得到该错误，但在test2（）的上例中.t._next_node总是包含垃圾值而不是nullptr我不明白为什么** < / p>

template<typename type>
LinkedList<type>& LinkedList<type>::operator=(const LinkedList<type>& other )
{
    if (&other == this)
        return *this;
    this->~LinkedList();
    this->_elem_nodes = nullptr;
    _size = 0;
    nodes<type> * cur_this_node = this->_elem_nodes;
    nodes<type> * cur_other_node = other._elem_nodes;
    while (cur_other_node != nullptr)
    {
        cur_this_node = new nodes<type>;
        cur_this_node->_value = cur_other_node->_value;
        this->_size++;
        cur_this_node = cur_this_node->_next_node;
        cur_other_node = cur_other_node->_next_node;
    }
    return *this;
}

template<typename type>
LinkedList<type>::LinkedList(const LinkedList<type>& other)
{
    _size = 0;
    nodes<type> * cur_this_node = this->_elem_nodes;
    cur_this_node = nullptr;
    nodes<type> * cur_other_node = other._elem_nodes;
    while (cur_other_node != nullptr)
    {
        cur_this_node = new nodes<type>;
        cur_this_node->_value = cur_other_node->_value;
        cur_this_node->_next_node = nullptr;
        this->_size++;
        cur_this_node = cur_this_node->_next_node;
        cur_other_node = cur_other_node->_next_node;
    }
}

Answer 1

您有三（或四或五）违规规则。您已经定义了自定义析构函数，但没有定义自定义赋值运算符。在您的情况下，这最终会导致两个单独的LinkedList对象指向相同的节点。

更多信息：https://stackoverflow.com/a/4782927/951890

自定义链接列表创建RtlValidateHeap错误，结构具有链接列表

1 个答案: