Implement Stack using Queues

push(s,  x)
1) Enqueue x to q1 (assuming size of q1 is unlimited).

pop(s)
1) One by one dequeue everything except the last element from q1 and enqueue to q2.
2) Dequeue the last item of q1, the dequeued item is result, store it.
3) Swap the names of q1 and q2
4) Return the item stored in step 2.
// Swapping of names is done to avoid one more movement of all elements from q2 to q1.
CODE

struct stack
{
  struct sNode *queue1;
  struct sNode *queue2;
};

void push(struct stack *s, int x)
{
  enqueue(&s-> queue1, x);
}

int pop(struct stack *s)
{
   int x; 
  
   if(s-> queue1== NULL && s->queue2== NULL)
   {
      printf("Stack is empty");
      exit(0);
   }

   while(queue1.getSize() != 1 )
   {
       x = queue1.dequeue();
       queue2.enqueue(x);
   }
   int value = queue1.dequeue();
   
   while( ! queue2.isEmpty() )
   {
       queue1.enqueue(queue2.dequeue() );
   }
   return value;
}

Stack  can also be implemented using one user Queue and one Function Call Queue (Recursion).
push(x)
1) Enqueue x to Queue1.

Pop():
1) If queue1 is empty then error.
2) If queue1 has only one element then return it.
3) Recursively pop everything from the queue1, store the popped item
in a variable res,  push the res back to queue1 and return res

struct stack
{
  struct sNode *queue1;
};

void push(struct stack *s, int x)
{
  enqueue(&s-> queue1, x);
}
  
int pop(struct stack *s)
{
   int x, res; 
  
   if(q->queue1== NULL)
   {
     printf("Stack is empty");
     exit(0);
   }
   else if(q->queue1->next == NULL)
   {
      return dequeue(&q->queue1);
   }
   else
   {
      x = dequeue (&q->queue1);
      res = pop(q);
      enqueue (&q->queue1, x);
  
      return res;
   }
}