How to Insert ,Delete ( first or last ) nodes in Linked List C / C++

Node definition (Structure) :

struct node{
     int num;
     node* next;
 };

The num variable stores the data given by the user and the pointer (next) stores the address of the next element present in the linked list.

Insertion of Nodes :

1) Inserting in beginning : 

void insert_beg(int value){
     node* n=new node;
     n->num=value;
     if(first==NULL){
         n->next=NULL;
         first=n;
     }
     else{
         n->next=first;
         first=n;
     }
 }

the above function ,when called creates a new node and stores the value passed to it in num (variable) ,now the function checks if the list is empty and if it is empty then it is simply marked as the first element in the linked list.

if the list is not empty then the pointer (next ) is assgned to the first element address and the marker (first) is given to current element.

2) Insertion at end : 

void insert_end(int value){
     node* n=new node;
     n->num=value;
     n->next=NULL;
     if(first==NULL)
         first=n;
     else{
         node* temp=first;
         while(temp->next!=NULL)
             temp=temp->next;
         temp->next=n;
     }
 }

This function is called when an element is to be stored in the last position of a linked list.

if the list is empty , we will add the element as first in the list.

if not then , we have to traverse the list first and stop when a node's (temp) next value is NULL and then point the node's (temp) next value to the node which is to be inserted.

Deletion of nodes :

1) Delete from beginning :

void delete_beg(){
     if(first==NULL){
         cout<<"Empty list\n";
     }   
     else{
         node* temp=first;
         first=first->next;
         delete(temp);
     }
 }

1) Check if the list is empty or not

2) if not then store the address of the top most ( first ) node in the list so that we can delete it later

3) point the first (tag) to the next element and delete the temp pointer pointing to the first element of the node.

2) Delete from ending :

void delete_end(){
     if(first==NULL)
         cout<<"Empty list\n";
     else{
         node *temp=first;
         while(temp->next->next)
               temp=temp->next;
         delete(temp->next);
         temp->next=NULL;
     }
 }

1) Check if the list is empty or not

2) if not then store the address of the top most ( first ) node in temp so that we can traverse the list until we reach the node which is second last node in the list ( where temp->next->next==NULL ),

here temp is the second last element in the list.

3) delete temp->next and point it to NULL value.

if any doubts or queries please comment