复制现有指针时出现 Seg 错误答案

【问题标题】：Seg fault error when copying an existing pointer复制现有指针时出现 Seg 错误
【发布时间】：2020-02-26 11:49:39
【问题描述】：

这是一个类项目的链表实现，除了这个功能外，它都可以工作。我已将 seg-fault 的位置缩小到指示的行（包含在 'std::cout...' statments 中。

// Purpose: removes a section of a lists
// Parameters:    start pointer to the begining of the section to remove.
//                end pointer to the end of the section to remove.
// Preconditions: start and end are pointers to nodes in this list.
//                start precedes end in this list.
// Postconditions: elements between start and end (inclusive) are removed
//                 from the list.
template<typename T>
void LinkedList<T>::clip(LLNode<T>* start, LLNode<T>* stop)
{
  if(m_size > 2)
  {
    LLNode<T>* walker = m_head;
    LLNode<T>* hold;
    LLNode<T>* deleter;
    while(walker != start)
    {
      hold = walker;
      walker = walker -> m_next;
    }
    int i;
    //TODO: seg-fault here
    if(stop == NULL)
    {
      hold -> m_next = NULL;
    }
    else if(stop -> m_next == NULL)
    {
      hold -> m_next = NULL;
    }
    else
    {
      std::cout << "error on next line\n";
      hold -> m_next = stop -> m_next;
      std::cout << "oh... i guess it's fixed...\n";
    }
    if(hold -> m_next == NULL)
    {
      m_back = hold;
    }
    stop -> m_next = NULL;
    walker = start;
    while(walker != NULL && walker -> m_next != NULL)
    {
      hold = walker;
      walker = walker -> m_next;
      delete deleter;
      deleter = hold;
      i++;
    }
    delete walker;
    delete deleter;
    m_size -= i;
  }
  else
  {
    clear();
  }
}

输出如下：

下一行出错

分段错误（核心转储）

任何帮助将不胜感激！

【问题讨论】：

当start 等于head 时，您不处理这种情况（那时hold 尚未初始化！）。
谢谢！我没有抓住那个！但是，问题仍然存在于列表中间的开始和结束。：/ 编辑：即使我的测试没有使用头部作为开始，它修复了它。我将发布更新的代码作为解决方案！谢谢！
你不需要单独处理 stop->m_next 和 nullptr ，这也包括在 else 后面。你应该更喜欢 C++ keywords (nullptr) 而不是旧的（过时的？）C macros (NULL)。
你可以更简单地删除：while(walker) { hold = walker; walker = walker->m_next; delete hold; ++i; } /*no deletes afterwards*/
if(m_size >/*!*/ 2) { } else { clear(); } - 如果stop 是start 的直接继任者或者 start 是第二个元素？你不想同时删除这两个元素......

标签： c++ linked-list segmentation-fault

【解决方案1】：

显然我忽略了“开始”等于头部的边缘情况。

这是完全固定的代码：

// Purpose: removes a section of a lists
// Parameters:    start pointer to the begining of the section to remove.
//                end pointer to the end of the section to remove.
// Preconditions: start and end are pointers to nodes in this list.
//                start precedes end in this list.
// Postconditions: elements between start and end (inclusive) are removed
//                 from the list.
template<typename T>
void LinkedList<T>::clip(LLNode<T>* start, LLNode<T>* stop)
{
  LLNode<T>* walker = m_head;
  LLNode<T>* deleter;
  LLNode<T>* lastNodeOnFront;
  if(m_head == start)
  {
    lastNodeOnFront = m_head;
  }
  while(walker != start)
  {
    lastNodeOnFront = walker;
    walker = walker -> m_next;
  }
  std::cout << *this;
  if(stop == m_back || stop == NULL)
  {
    lastNodeOnFront -> m_next = NULL;
  }
  else
  {
    lastNodeOnFront -> m_next = stop -> m_next;
  }
  std::cout << "before delete\n" << *this;
  walker = start;
  while(walker != lastNodeOnFront -> m_next)
  {
    deleter = walker;
    walker = walker -> m_next;
    delete deleter;
    m_size--;
  }
  return;
}

感谢阿空加瓜的帮助！

【讨论】：