Adding/ removing tail, showing list in reverse. linked list c++ - c++

Question

I am working on a linked list, I made a menu (As a noob this greatly impressed me:)) and got the add front and remove front to work. although I want to be able to add in a node at the end then remove it and display it in reverse. I thought my removelast function looks like it would work but it doesn't. Not sure how to add to the tail, and no clue what so ever how to show in reverse. I looked up some guides but i am no good at understanding from text, I like to see examples pertaining to my problem.
Take a look!
#include <iostream>
using namespace std;
struct node //node structure
{
int number;
node *next;
};
bool isempty(node *head);
char menu();
void first(node *&head, node *&last, int number );
void insert(node *&head, node *&last, int number);
void removefirst(node *&head, node *&last);
void removelast(node *&head, node *&last);
void shownode(node *current);
void showreverse(node *&last);
bool isempty(node *head)
{
if(head == NULL)
return true;
else
return false;
}
char menu()
// I added a menu because in an actual application this is what you would have
//And it was fun :D
{
char choice;
cout<<"choose an option:"<<endl;
cout<<"1. Add Node."<<endl;
cout<<"2. Add Node to Last"<<endl
cout<<"3. Remove First Node."<<endl;
cout<<"4. Remove last node"<<endl;
cout<<"5. Show Node List."<<endl;
cout<<"6. Reverse Node List."<<endl;
cout<<endl;
cin>>choice;
cout<<endl;
return choice;
}
void first(node *&head, node *&last, int number ) //adding first
{
node *temp =new node;
temp->number = number;
temp->next = NULL;
head = temp;
last = temp;
}
void insert(node *&head, node *&last, int number)//adding more
{
if(isempty(head))
first(head, last, number);
else
{
node *temp =new node;
temp->number = number;
temp->next = NULL;
last->next = temp;
last = temp;
}
}
void addlast (node *&head, node *&last, int number)
{
node *temp =new node;
temp->number = number;
temp->last = NULL;
last->next = temp;
last = temp;
}
void removefirst(node *&head, node *&last)//destructor
{
if(isempty(head))
cout<<"List is empty."<<endl;
else if (head == last)
{
delete head;
head == NULL;
last == NULL;
}
else
{
node *temp = head;
head = head->next;
delete temp;
}
}
void removelast(node *&head, node *&last)
{
if(isempty(head))
cout<<"List is empty."<<endl;
else if (head == last)
{
delete last;
head == NULL;
last == NULL;
}
else
{
node *temp = last;
last = last->next;
delete temp;
}
}
void shownode(node *current)
{
if (isempty(current))
cout<<"list is empty"<<endl;
else
{
cout<<"Nodes in list:"<<endl;
while(current != NULL)
{
cout<<current->number<<endl;
current = current->next;
}
}
}
void showreverse(*&last)
{
}
int main()
{
node *head = NULL;
node *last = NULL;
char choice;
int number;
do{
choice = menu();
switch(choice)
{
case '1': cout<<"inert number:"<<endl;
cin>>number;
insert(head, last, number);
break;
case '2': addlast(head, last, number);
break;
case '3': removefirst(head, last);
break;
case '4': removelast(head, last);
break;
case '5': shownode(head);
break;
case '6': showreverse(last);
break;
default: cout<<"Exit";
}
}while(choice != '4');
return 0;
}

Answer 1

In your removelast(node *&head, node *&last) function when you delete the last element, the next member of the last but one node will still point to the last element you deleted.
node *temp = last;
last = last->next;
delete temp;
You are making a pointer to the same node, change its next member and than delete it, so the first two instructions are useless. The problem is at the last but one next member, but you have not way of changing it without iterating over the whole list. I think the best approach is to use a doubly linked list. With doubly linked list just interchange last pointer with head pointer, or parse it from last to head.