Circular linked list C++ - c++

Question

I'm trying to create a queue using a circular linked list. I just took my existing linked list implementation (which already works) and changed where the nodes pointed but it doesn't seem to be that straightforward. I'm sure it's one small thing that I'm overlooking.
void enqueue(int q){
newNode=new node;
newNode->info = q;
newNode->link = first;
if(first->link==first){
first = last = newNode;
last->link=first;
}
else{
last->link = newNode;
last= newNode;
last->link = first;
}
}
int dequeue(){
int x;
if(first->link==first){
cout <<"The Queue is empty" <<endl;
return 0;
}
else{
dummy=first;
x=dummy->info;
first=first->link;
delete(dummy);
return(x);
}
}

Answer 1

I converted this to a class. For a circular list class, only a pointer to last node is needed, as first = last->next. Since no dummy node is used, an empty list is normally indicated by last == NULL;
#include<iostream>
class circq
{
typedef struct node_
{
struct node_ *next;
int info;
}node;
node *last; // ptr to last node of list
public:
circq() // constructor
{
last = NULL; // reset last
}
bool isempty()
{
return last == NULL;
}
void enqueue(int q)
{
node *newNode=new node; // new node
newNode->info = q;
if(last == NULL){ // if empty list
last = newNode; // create single node list
newNode->next = newNode;
return;
} // else
newNode->next = last->next; // append node to end list
last->next = newNode;
last = newNode;
}
int dequeue()
{
if(last == NULL){ // check for empty list
return 0;
}
node *first = last->next; // set ptr to first node
int x = first->info; // set return value
if(last->next == last) // if single node
last = NULL; // set list to empty
else // else
last->next = first->next; // advance first to next node
delete first; // delete node
return x;
}
~circq() // destructor
{
while(!isempty())
dequeue();
}
};
int main() // test circq
{
circq cq;
cq.enqueue(0);
cq.enqueue(1);
cq.enqueue(2);
cq.enqueue(3);
while(!cq.isempty())
std::cout << cq.dequeue() << std::endl;
return 0;
}