具有嵌套类作为私有类的类中具有析构函数的内存泄漏

Question

我有一个Tree类，它有一个嵌套的私有Node类。我的[[Tree类的构造函数创建一个Node对象，并将其设置为_info，该对象是我的[[Tree类中的Node类型的指针。我没有任何错误，但是当我使用valgrind检查程序是否存在内存泄漏时，我发现我存在内存泄漏。我知道问题在于解构函数，但我不知道如何解决。在显示代码之前，我应该提到模板的第二个参数（N）在这里并不重要，您可以忽略它。

这是我的代码：

---

tree.hpp

#ifndef TREE_HPP
#define TREE_HPP

#include <iostream>
#include <cstddef> 

template <typename T, char N>
class Tree {
private:
  class Node;
  Node* _info;
public:
  Tree();
  Tree(T, char);
  Tree(const Tree&) = delete;
  Tree& operator= (const Tree&) = delete;
  Tree (Tree&&);
  Tree& operator= (Tree&&);
  ~Tree() {delete _info;} // something is wrong here
  bool ins(char, Tree*);
  Tree* fils(char);
  void print(){
    std::cout << _info->getData() << std::endl;
  }
};

template <typename T, char N>
bool Tree<T, N>::ins(char index, Tree* childTree){
  if (_info){
    _info->getChildren()[index] = childTree;
    return true;
  }
  return false;
}

template <typename T, char N>
Tree<T, N>::Tree() 
  : _info(nullptr) {}

template <typename T, char N>
Tree<T, N>::Tree(T data, char size) {
  Node* node = new Node(data, size); // I think I don't free this
  _info = node;
}

template <typename T, char N>
Tree<T, N>::Tree(Tree&& t) {
  _info = t._info;
  t._info = nullptr;
}

template <typename T, char N>
Tree<T, N>& Tree<T, N>::operator= (Tree&& t) {
  if (&t != this) {delete _info; _info = t._info; t._info = nullptr;}
  return *this;
}

template <typename T, char N>
typename Tree<T,N>::Node* Tree<T, N>::info() { return _info;}

template <typename T, char N>
Tree<T,N>* Tree<T, N>::fils(char index){
  return _info->getChildren()[index];
}

template <typename T, char N>
class Tree<T, N>::Node {
private:
  T _data;
  Tree* _children;
  bool _isWord;
public:
  Node();
  Node(T, char);
  Tree** getChildren() {return &this->_children;}
  T getData(){return this->_data;}
  ~Node() = default;
};

template <typename T, char N>
Tree<T,N>::Node::Node(){
  _data = 0;
  _children = nullptr;
  _isWord = false;
}

template <typename T, char N>
Tree<T,N>::Node::Node(T data, char size){
  _data = data;
  _isWord = false;
  _children = new Tree<T,N>[size]; 
}

#endif

main.cpp

#include <iostream>
#include <cstddef>
#include <exception>
#include "tree.hpp"

#define SIZE 5
int main() {
  Tree<char,SIZE> n1('A',SIZE);
  Tree<char,SIZE> n1_1('B',SIZE);
  Tree<char,SIZE> n1_2('C',10); // ! here the size is always 5
  Tree<char,SIZE> n1_1_1('D',SIZE);
  n1.ins(0,&n1_1);
  n1.ins(1,&n1_2);
  n1_1.ins(0,&n1_1_1);
  n1.fils(0)->print();
  n1.fils(1)->print();  
  n1_1.fils(0)->print();
  return 0;
}

我也愿意提出任何改进我的代码的建议。谢谢高级

Answer 1

_children = new Tree<T,N>[size];

[无处没有delete[] _children;。

甚至有更多问题：
Tree** getChildren() {return &this->_children;}

此函数返回一个指向不是数组元素的单独Tree*对象的指针。

return _info->getChildren()[index];

这里，您将索引指向指向孤立对象的指针，就好像它在数组中一样。使用0以外的任何其他索引都有UB。