【发布时间】:2014-09-28 20:44:01
【问题描述】:
我有一个任务是编写一个反转函数来反转一个最多包含 n 个块的双向链表。我首先从 forloop 中的大小获取端点,然后将 tartpoint 和端点发送到外部函数以反转它们。该外部函数成功地反转了头部和尾部,但是我在尝试反转给定大小时出现了段错误。我需要一些关于出了什么问题的帮助?
反向函数;
/**
* Helper function to reverse a sequence of linked memory inside a List,
* starting at startPoint and ending at endPoint. You are responsible for
* updating startPoint and endPoint to point to the new starting and ending
* points of the rearranged sequence of linked memory in question.
*
* @param startPoint A pointer reference to the first node in the sequence
* to be reversed.
* @param endPoint A pointer reference to the last node in the sequence to
* be reversed.
*/
template <class T>
void List<T>::reverse( ListNode * & startPoint, ListNode * & endPoint )
{
if( (startPoint == NULL) || (endPoint == NULL) || (startPoint == endPoint))
{ return; }
ListNode * curr = startPoint;
ListNode * nexter = NULL;
ListNode * prever = endPoint->next;
while( curr != NULL)
{
nexter = curr->next;
curr->next = prever;
prever = curr;
curr = nexter;
prever->prev = curr;
}
// now swap start and end pts
nexter = startPoint;
startPoint = endPoint;
endPoint = nexter;
}
现在给定sze的反向函数,应该使用上面的函数;
/**
* Reverses blocks of size n in the current List. You should use your
* reverse( ListNode * &, ListNode * & ) helper function in this method!
*
* @param n The size of the blocks in the List to be reversed.
*/
template <class T>
void List<T>::reverseNth( int n )
{
if(n == 0)
return;
ListNode * startPoint = head;
ListNode * endPoint = head;
ListNode * save = NULL;
for(int i = 0; i< n; i++) // need to get endpoint at n
{
endPoint = endPoint->next;
}
reverse(startPoint, endPoint);
}
gdb 输出一些奇怪的东西,可能是因为之后处理图像的函数失败了;
Program received signal SIGINT, Interrupt.
0x000000000040dcab in __distance<List<RGBAPixel>::ListIterator> (__first=..., __last=...) at /class/cs225/llvm/include/c++/v1/iterator:488
488 for (; __first != __last; ++__first)
(gdb) q
A debugging session is active.
Inferior 1 [process 31022] will be killed.
【问题讨论】:
-
endPoint = endPoint->next;如果你到达列表的末尾,你最终会取消引用一个空指针。 -
您是否尝试查看回溯 (
bt)? -
我输入了一段时间(endpoint-> != NULL),但程序仍然进入无限循环。回溯将我带到我们不应该接触的其他功能
-
由于您的列表是双向链接的,因此您需要做的就是在列表中遍历每个节点的
next和prev字段中的值。然后通过交换startPoint和endPoint来完成。 -
这就是我在顶级函数中所做的
标签: c++ function pointers linked-list segmentation-fault