removing two items from a stack and adding their values - stack

Question

I am writing a program that creates a stack using linked lists. I have all the functionality finished like push(), pop(), top(), etc.
The thing I am trying to figure out is how to remove two values from the stack, add them together, then push that into the stack. This is part of the assignment and we have to continue to do it until all the items are added together and only the sum remains in the stack.
Any help or tips would be appreciated!
EDIT: I solved my problem by just making another function!
Thank you to everyone who tried to help!
Here is my code:
#include <cstdlib>
#include <iostream>
using namespace std;
//Creating a node structure
struct node
{
int data;
struct node *next;
};
//class stack
class stack
{
struct node *top;
int size;
public:
stack()
{
top=NULL;
}
void push(); // insert an element
void pop(); // delete an element
void stackTop(); //retrive top item without removal
void stackSize(); //return the size of the stack
void isEmpty(); // return 1 if empty, 0 if not
void show(); // show the stack
};
//push items into a stack
void stack::push()
{
int value;
struct node *ptr;
cout << "\nEnter a number to insert: ";
cin >> value;
ptr = new node;
ptr->data = value;
ptr->next = NULL;
if(top != NULL)
{
ptr->next = top;
}
top = ptr;
cout<<"\nNew item is inserted to the stack!!!" << endl;
size ++;
}
//remove the top item
void stack::pop()
{
struct node *temp;
if(top == NULL)
{
cout<<"\nThe stack is empty!!!";
return;
}
temp = top;
top = top->next;
cout << "\nPoped value is " << temp->data << endl;
delete temp;
size--;
}
//retrive top value without removing it
void stack::stackTop()
{
struct node *temp;
if(top == NULL)
{
cout<<"\nThe stack is empty!!!";
return;
}
temp = top;
cout << "The top item is: " << temp->data << endl;
delete temp;
}
//show the stack
void stack::show()
{
struct node *ptr1 = top;
cout << "\nThe stack is:" << endl;
while(ptr1 != NULL)
{
cout << ptr1->data << " ->";
ptr1 = ptr1->next;
}
cout << "NULL" << endl;
}
//return if empty or not
void stack::isEmpty()
{
if(top == NULL)
{
cout<<"\nThe stack is empty!!!" << endl;
return;
}
else
{
cout << "\nThe stack is not empty!!!" << endl;
return;
}
}
//return the number of items in the stack
void stack::stackSize()
{
if(top == NULL)
{
cout<<"\nThe stack is empty!!!" << endl;
return;
}
else
{
cout << "\nThe stack has " << size << " items" << endl;
return;
}
}
//main function
int main()
{
stack s;
int choice;
while(1)
{
cout << "\nSTACK USING LINKED LIST" << endl << endl;
cout << "1:PUSH" << endl;
cout << "2:POP" << endl;
cout << "3:DISPLAY STACK" << endl;
cout << "4:RETRIVE TOP ITEM" << endl;
cout << "5:GET THE SIZE" << endl;
cout << "6:IS THE STACK EMPTY?" << endl;
cout << "7:EXIT" << endl;
cout << "Enter your choice(1-7): ";
cin >> choice;
switch(choice)
{
case 1:
s.push();
break;
case 2:
s.pop();
break;
case 3:
s.show();
break;
case 4:
s.stackTop();
break;
case 5:
s.stackSize();
break;
case 6:
s.isEmpty();
break;
case 7:
exit(0);
break;
default:
cout << "\nPlease enter correct choice(1-7)!!!" << endl;
break;
}
}
return 0;
}

Answer 1

Is the interface you have mandatory? Normally you would have your pop() operation return the value you just popped (rather than void), instead of just printing it. If you do that, your problem becomes simple, and you can just repeatedly use your algorithm to add them together.
As a matter of fact, pop(), stackTop(), stackSize(), and isEmpty() should probably all return their values. If the print statements aren't required within the functions, you could then just have your main program print the values it finds.