C++中链表的错误"Abort signal from abort(3) (sigabrt) "

Question

以下代码用于基本的循环链表，但是当输入n(例如8位数字(的大值时，它会抛出"abort(3(中止信号(sigabrt("错误。我不确定这意味着什么，并希望获得一些有关修复我的代码的指导。 谢谢！

#include<bits/stdc++.h> 
using namespace std; 
//First I created a structure for a node in a circular linked list
struct Node 
{ 
int data; 
struct Node *next; 
}; 
// function to create a new node
Node *newNode(int data) 
{ 
Node *temporary = new Node; 
temporary->next = temporary; 
temporary->data = data; 
return temporary; 
} 
// This function finds the last man standing in
//the game of elimination
void gameOfElimination(int m, int n) 
{ 
//first I created a circular linked list of the size which the user inputted 
Node *head = newNode(1); 
Node *prev = head; 
//this loop links the previous node to the next node, and so on.
for (int index = 2; index <= n; index++) 
{ 
prev->next = newNode(index); 
prev = prev->next; 
} 
prev->next = head; //This connects the last and first nodes in our linked list together. 
//when only one node is left, which is our answer:
Node *ptr1 = head, *ptr2 = head; 
while (ptr1->next != ptr1) 
{ 
int count = 1; 
while (count != m) 
{ 
ptr2 = ptr1; 
ptr1 = ptr1->next; 
count++; 
} 
/* Remove the m-th node */
ptr2->next = ptr1->next; 
ptr1 = ptr2->next; 
} 
printf ("%dn ", 
ptr1->data);
} 
//main program which takes in values and calls the function:
int main() 
{ 
int n, p;
cin>>n>>p;
int m=p+1;
gameOfElimination(m, n); 
return 0; 
} 

Answer 1

SIGABRT通常在存在内存问题时发出(堆损坏很常见(。在您的代码中，我只看到正在调用的 new(( 运算符，但您没有从链表中删除任何未使用的节点！似乎您正在耗尽分配给进程的内存。

Answer 2

您可能内存不足。在程序执行期间检查您的 ram 使用情况，这可能会导致某些事情。

Answer 3

enter code here

#include<bits/stdc++.h>
using namespace std;
class Node{
public: 
int data;
Node *next;
};
void traverse(Node *head)
{
while (head != NULL)
{
/* code */
cout<<head->data<<"->";
head = head->next;
}
cout<<"NULL"
}
int main()
{
Node *head = new Node();
Node *second = new Node();;
Node *third = new Node();;
Node *fourth = new Node();;

head->data = 5;
head->next = second;
//cout<<head->data;
second->data=10;
second->next=third;
third->data = 15;
third->next = fourth;
fourth->data = 20;
fourth->next= NULL;
traverse(head);
return 0;
}```