C++ linked list nodes - c++

Question

I have very basic concepts of linked list with C++. Here I have nodes linked, but the idea is to delete the last node, how do I achieve this?
This is the portion of code aimed to delete the last node:
//deleting node here
age* temp = head;
temp->next->next;//!=NULL
temp = temp->next;
//temp->next = NULL;
delete temp;
#include<iostream>
using namespace std;
struct age{
int a;
age *next;
};
age *head,*current,*node1,*node2,*ona;
int main(){
//creating a first node
age *node1=new age();
head=node1;
node1->a=10;
//creating a second node
age *node2=new age();
node2->a=20;
//link nodes
node1->next=node2;
node2->next=NULL;
//insertion of node ona between node 1 and node 2
ona=new age;
ona->a=15;
ona->next=node1->next;
node1->next=ona;
//deleting node here
age* temp = head;
temp->next->next;//!=NULL
temp = temp->next;
//temp->next = NULL;
delete temp;
//displaying the otput
current=head;
while(current!=NULL){
cout<<current->a<<endl;
current=current->next;
}
}

Answer 1

I suggest to have a look here:
For plain C development: https://www.learn-c.org/en/Linked_lists
On this site all standard methods for handling linked lists are explained and you can find code snippets for every operation.
For CPP development: https://www.codesdope.com/blog/article/c-deletion-of-a-given-node-from-a-linked-list-in-c/
On this site you can find an example coded in CPP OOP style.
I change the C example a little bit to fit your code example:
void remove_last(age * head) {
/* if there is only one item in the list, remove it */
if (head->next == NULL) {
delete head;
head = NULL;
return;
}
/* get to the second to last node in the list */
node_t * current = head;
while (current->next->next != NULL) {
current = current->next;
}
/* now current points to the second to last item of the list, so let's remove current->next */
delete(current->next);
current->next = NULL;
}