C++ 代码在运行时挂起

代码在执行时挂起。请帮助我了解这里的问题是什么 它在第一次遍历后进行STCK，甚至不进入反向功能。无法理解此问题： 帮我吧

# include<iostream>
# include<stdlib.h>

using namespace std;

struct sLL
{
   int data;
   sLL * next;
};

void createList(sLL ** head, int n)
{
   if(n == 0)
   {
      return;
   }
   sLL * temp = (sLL *) malloc(sizeof(sLL));
   if(temp != NULL)
   {
      temp->data = rand() % 100;
      temp->next = NULL;
      *head = temp;
      //head = &(temp->next);
      createList(&(temp->next), n - 1);
   }
}

void traverse(sLL * head)
{
   cout<<"In traverse"<<endl;
   while(head) {
      cout<<head->data<<"->";
      head=head->next;
   }
   cout<<"NULL"<<endl;
   cout.flush();
   return;
} 

 sLL* reverse(sLL * head)
 {
    sLL * temp = NULL, * newNode = NULL;
    while(head) {   
       newNode = head->next; // to traverse forword
       head->next = temp;
       temp = head; // Current node value which we use in next itr as a previous node value.
       newNode = head; // assigning newNode vlaue back to head so that we can traverse forward.
    }
    return temp; // final node value will be in temp i.e the current node value.
}

int main()
{
    int n;
    sLL * head = NULL, *temp = NULL;
    cout<<"Enter the number of nodes :";
    cin>>n;
    createList(&head, n);
    cout<<"\nCreate Done"<<endl;
    traverse(head);
    cout.flush();
    cout<<"Reverse Start";
    temp = reverse(head);
    cout<<"reverse done";
    traverse(temp);
}

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您根本不需要重新分配头值

while(head)
    //loop

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在这里，头指针永远不会改变。因此循环是无限的。

您的评论与此处的代码相矛盾：

newNode = head; // assigning newNode vlaue back to head so that we can traverse forward.

如果你按照评论所说的去做，应该会更好：

head = newNode; // assigning newNode value back to head so that we can traverse forward.

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您能指出横移函数SIR head=head->next的while循环中的@saurabh的位置吗；是否存在…@saurabh抱歉，反向函数不是无限循环。。。。。在我旅行的同时，旅行也进行得很好，超过了打印的值吗？通过有选择地注释，直到只剩下很少的行，缩小问题的范围。这是呼叫调试，可能会为您解决问题…明白了，谢谢aswin已经解决了