自定义链表创建 RtlValidateHeap 错误,其结构具有链表

custom linked list creating RtlValidateHeap error with the structures having a Linked list

本文关键字:链表 结构 错误 创建 RtlValidateHeap 自定义      更新时间:2023-10-16

这是我正在做的LinkedList实现。 它适用于任何数据类型,但是当我尝试制作具有链表的类型的链表时会出现问题 关于使用Visual Studio进行调试,我得到了 指定给 RtlValidateHeap 的地址无效( 00790000, 007B16D0 )

代码如下所示:

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 );
}

下面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,但不再收到该错误,但在test2()的上述示例中.t._next_node总是包含垃圾值而不是nullptr,我不明白为什么**

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;
}
}

您有三次(或四次)违规的规则。 你已定义自定义析构函数,但尚未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