How to use point to next node when deleting elements in C++ - c++

Question

I have previously posted some part of this task here.
I am now implementing a method that removes an element at a given index. My code is
void remove(int index)
{
if (head != NULL)
{
Node *current = get_node(index);
Node *prev = get_node(index - 1);
Node *next = get_node(index + 1);
prev->next = current->next;
delete current;
}
}
however, I am facing this error message
libc++abi.dylib: terminating with uncaught exception of type
std::range_error: IndexError: Index out of range
Abort trap: 6
I am guessing the problem is the pointers, but I am not sure why this is not working. Anyone who can help?

Answer 1

I think you can handle corner cases like this:
#include <iostream>
using namespace std;
struct Node {
Node(int val) {
this->val = val;
}
struct Node * next;
int val;
};
class LinkedList {
public:
Node* head;
LinkedList() {
head = new Node(1);
Node * n1 = new Node(2);
head->next = n1;
Node * n2 = new Node(3);
n1->next = n2;
}
void remove(int index) {
if (head == NULL) {
return;
}
int pos = 0;
Node * cur = head;
Node *prev = NULL;
while (cur != NULL) {
if (pos == index) {
break;
}
pos++;
prev = cur;
cur = cur->next;
}
if (prev == NULL) {
head = head->next;
}
else {
prev->next = cur->next;
}
delete cur;
}
};
void print(Node * head){
cout << "Current linked list:\n";
while(head != NULL) {
cout << head->val << endl;
head = head->next;
}
cout << endl;
}
int main() {
LinkedList * list = new LinkedList();
print(list->head);
list->remove(0);
print(list->head);
list->remove(1);
print(list->head);
list->remove(0);
print(list->head);
}