从解除引用的迭代器返回std :: map <std :: string,int>时发生大量内存泄漏

Question

**小心**,这个程序不仅会挂起,而且显然会永远占用你的所有记忆,让你的计算机变得缓慢而糟糕.很长一段时间以来,我一直在努力解决这个问题,并且想出了很多东西 - 除了为什么它真的挂了.很抱歉,有太多代码,但我删除了所有不相关的内容,这就是剩下的内容.

链表

//=====================
// Linked List

#include <stdexcept>

template<class T> struct LinkedList {
    public:
        LinkedList();
        LinkedList(const LinkedList& srcList);
        ~LinkedList();

        void addObject (T& addedObject);

        class ListIterator {
            public:
                ListIterator();
                explicit ListIterator(LinkedList<T>& parentList);

                // Operators
                ListIterator& operator++();
                T& operator*() const;
                bool operator!=(const ListIterator& otherIter);

            private:
                typename LinkedList::Node* current_;
        };

        ListIterator begin();
        ListIterator end();
        std::size_t size_;

    private:
        struct Node {
            Node();
            Node(T& object);
            Node(const Node&) = delete;
            T* const object_;
            Node* next_;
            Node* prev_;
        };

        Node head_;
        Node tail_;
};

//====================
// Classes (Implementation)

// Linked List default constructor 
template<class T> LinkedList<T>::LinkedList() 
: size_{0} {
    head_.next_ = &tail_;
    tail_.prev_ = &head_;
};

// Linked List copy constructor
template<class T> LinkedList<T>::
LinkedList(const LinkedList& srcList) { 
    size_ = srcList.size_;
    head_.next_ = &tail_;
    tail_.prev_ = &head_;
    ListIterator nodesToCopy = srcList.begin();

    while (nodesToCopy != srcList.end()) { 
        this->addObject(*nodesToCopy);
        srcList.removeObject(1);
    };
    delete &srcList;
};

// Linked List destructor
template<class T> LinkedList<T>::~LinkedList() {
    for (unsigned int ii = 1; ii == size_; ++ii) {
        Node* toDelete = head_.next_;
        head_.next_ = head_.next_->next_;
        delete toDelete;
    };
};

// Add object to Linked List
template<class T> void LinkedList<T>::addObject(T& addedObject) {
    Node* node = new Node(addedObject);
    node->prev_ = tail_.prev_;
    tail_.prev_->next_ = node;
    tail_.prev_ = node;
    node->next_ = &tail_;
    ++size_;
};

// Linked List Iterator constructor
template<class T> LinkedList<T>::ListIterator::
ListIterator(LinkedList<T>& parentList) {
    current_ = parentList.head_.next_;
};

// Iterator operators
// Increment forward
template<class T> typename LinkedList<T>::ListIterator& LinkedList<T>::
ListIterator::operator++() {
    current_ = current_->next_;
    return *this;
};

// Return object pointed to
template<class T> T& LinkedList<T>::ListIterator::
operator*() const {
    return *(current_->object_);  
};

template<class T> bool LinkedList<T>::ListIterator::
operator!=(const ListIterator& otherIter) { 
    return &(**this) != &(*otherIter);
};

// Return an iterator object via begin() and end()
template<class T> typename LinkedList<T>::ListIterator
LinkedList<T>::begin() {
    ListIterator beginIterator(*this);
    return beginIterator;
};
template<class T> typename LinkedList<T>::ListIterator
LinkedList<T>::end() {
    ListIterator endIterator(*this);
    for (unsigned int ii = 0; ii < size_; ++ii) { ++endIterator; }; 
    return endIterator;
};

// Node constructors
template<class T> LinkedList<T>::Node::Node()
: object_(nullptr), next_(nullptr), prev_(nullptr) {};

template<class T> LinkedList<T>::Node::Node(T& object) 
: object_(&object) {};

项目

//=====================
// Item
//====================
// Included dependencies
#include <string>
#include <array>
#include <map>
#include <iostream>

class Item {
    public: 
        Item();
        Item(std::string name);
        Item(std::string name, std::array<int, 2> stats);
        std::map<std::string, int> getStats();

        std::string name_;

    private:
        std::map<std::string, int> enhancements_;
};

// Constructors
Item::Item() { 
    enhancements_["Str"] = 0;
    enhancements_["Def"] = 0;
};

Item::Item(std::string name) : Item::Item() { name_ = name; };
Item::Item(std::string name, std::array<int, 2> stats)
: Item::Item(name) {
    enhancements_["Str"] = stats[0];
    enhancements_["Def"] = stats[1];
};

// Return map of stats
std::map<std::string, int> Item::getStats() { return enhancements_; };

房间

//====================
// Room
class Room {
    public:
        void addItem(Item item);
        LinkedList<Item>::ListIterator getItems();
        LinkedList<Item> itemsInThisRoom_;
};

// Add item to room
void Room::addItem(Item item) { itemsInThisRoom_.addObject(item); };

// Get iterator which iterates over items in room
LinkedList<Item>::ListIterator Room::getItems() { 
    return itemsInThisRoom_.begin(); 
};

主要

int main() {
    std::array<int, 2> swordStats = {{5, 0}};
    std::array<int, 2> shieldStats = {{0, 2}};
    std::array<int, 2> armorStats = {{0, 3}};

    Item sword("Sword", swordStats);
    Item shield("Shield", shieldStats);
    Item armor("Armor", armorStats);
    Room room;

    room.addItem(shield);
    room.addItem(sword);
    room.addItem(armor);
    LinkedList<Item>::ListIterator roomItems = room.itemsInThisRoom_.begin();

    while (roomItems != room.itemsInThisRoom_.end()) {
        (*roomItems).getStats();
        ++roomItems;
    };

    return 0;
};

所有这些都可以放在一个文件中并进行编译(我按类将其拆分以使其更易于阅读).这是主要的,它挂起的线:

(*roomItems).getStats();

这让我相信我的解引用运算符有问题,对吧？如果我们在Room类之外创建一个迭代器,取消引用它,并以相同的方式获取getStats - 一切正常.

...所以这是Room课程的问题？

但是,如果我们将Item和main更改为以下内容:

//=====================
// Item
//====================
// Included dependencies
#include <string>
#include <array>
#include <map>
#include <iostream>

class Item {
    public: 
        Item();
        Item(std::string name);
        Item(std::string, int);
        int getStats();

        std::string name_;

    private:
        int enhancements_;
};

// Constructors
Item::Item() { 
    enhancements_ = 0;
};

Item::Item(std::string name) : Item::Item() { name_ = name; };
Item::Item(std::string name, int stats)
: Item::Item(name) {
    enhancements_ = stats;
};

// Return map of stats
int Item::getStats() { return enhancements_; };

//====================
// Room
class Room {
    public:
        void addItem(Item item);
        LinkedList<Item>::ListIterator getItems();
        LinkedList<Item> itemsInThisRoom_;
};

// Add item to room
void Room::addItem(Item item) { itemsInThisRoom_.addObject(item); };

// Get iterator which iterates over items in room
LinkedList<Item>::ListIterator Room::getItems() { 
    return itemsInThisRoom_.begin(); 
};

int main() {
    Item sword("Sword", 1);
    Item shield("Shield", 2);
    Item armor("Armor", 3);
    Room room;

    room.addItem(shield);
    room.addItem(sword);
    room.addItem(armor);
    LinkedList<Item>::ListIterator roomItems = room.itemsInThisRoom_.begin();
    while (roomItems != room.itemsInThisRoom_.end()) {
        (*roomItems).getStats();
        ++roomItems;
    };

    return 0;
};

一切都运行得很好.我可以返回int值.

...所以...它既不是Room类或dereference运算符的问题,而是返回 std :: map？GDB没有太多话要说.当我在违规线和步骤中突破时,我得到:

24  std::map<std::string, int> Item::getStats() { return enhancements_; };
(gdb) step
_Rb_tree_impl (__a=<optimized out>, __comp=..., this=0x7fffffffced0)
    at /usr/include/c++/4.9/bits/stl_tree.h:474
474         _M_header(), _M_node_count(0)
(gdb) step
475       { _M_initialize(); }
(gdb) step
_M_initialize (this=0x7fffffffced0)
    at /usr/include/c++/4.9/bits/stl_tree.h:484
484         this->_M_header._M_left = &this->_M_header;
(gdb) step
485         this->_M_header._M_right = &this->_M_header;
(gdb) step
_Rb_tree (__x=..., this=0x7fffffffced0)
    at /usr/include/c++/4.9/bits/stl_tree.h:674
674     if (__x._M_root() != 0)
(gdb) step
_M_root (this=0x7fffffffd048)
    at /usr/include/c++/4.9/bits/stl_tree.h:498
498       { return this->_M_impl._M_header._M_parent; }
(gdb) step
_Rb_tree (__x=..., this=0x7fffffffced0)
    at /usr/include/c++/4.9/bits/stl_tree.h:674
674     if (__x._M_root() != 0)
(gdb) step
676         _M_root() = _M_copy(__x._M_begin(), _M_end());
(gdb) step
std::_Rb_tree<std::string, std::pair<std::string const, int>, std::_Select1st<std::pair<std::string const, int> >, std::less<std::string>, std::allocator<std::pair<std::string const, int> > >::_M_copy (
    this=this@entry=0x7fffffffced0, __x=0x619f10, 
    __p=__p@entry=0x7fffffffced8)
    at /usr/include/c++/4.9/bits/stl_tree.h:1207
1207          _Link_type __top = _M_clone_node(__x);

......这对我来说是胡言乱语.:(它无限地这样做,所以我知道它(不知何故)描述了挂断.

我不知道这里发生了什么,哈哈.我对C++很陌生,自从我醒来以后就一直在努力解决这个问题,所以我知道我的代码很糟糕,我应该感觉不好写.

有任何想法吗？

Answer 1

除了已经提到的内容之外,您的Node对象还会无关地存储指针,该指针是通过引用从外部传递的对象

template<class T> LinkedList<T>::Node::Node(T& object) 
: object_(&object) {};

但是,传递给构造函数的引用参数Node实际上绑定到局部变量

template<class T> void LinkedList<T>::addObject(T& addedObject) {
    Node* node = new Node(addedObject);
    node->prev_ = tail_.prev_;
    tail_.prev_->next_ = node;
    tail_.prev_ = node;
    node->next_ = &tail_;
    ++size_;
};

void Room::addItem(Item item) { itemsInThisRoom_.addObject(item); };

即引用绑定到参数item,该参数是内部的局部变量addItem.

item一旦addItem退出,该局部变量就会被销毁.你的Node::object_指针仍然没有指向.

考虑到您在代码中执行的无偿复制的数量,完全不清楚您是如何设法想出存储指向您内部的非拥有对象的指针Node(而不是无偿地将整个数据复制到Node您的代码中)几乎在其他地方做.)

无论如何,代码中的内存所有权完全被破坏,导致上面的对象生命周期问题.您需要从头开始设计一些有意义的内存所有权计划,然后按照该计划编写代码.你现在拥有的是不可挽救的混乱.

如果你想使用指针并且你不认为你已经准备好解开这个混乱,只需使用智能指针让他们为你处理事情.

PS并放弃了;每一个之后放置一个令人讨厌的习惯}.