Double Pointers Queue, deQueue, enQueue

Question

Currently preparing for exams and can't figure out the reasoning behind the queue[rear*] = new_node in this example code.

are you not putting the address of new_node in rather than the value?

Also, is the queue a Node ** queue because it is a pointer to a list of node pointers?

Thanks so much, I really appreciate it, no matter how many hours i spend on double pointers, they always crop up and retest my understanding i thought i finally had!

void enQueue(struct node **queue, int *rear, struct node *new_node) 
{ 
queue[*rear] = new_node; 
(*rear)++; 
} 

struct node *deQueue(struct node **queue, int *front) 
{ 
(*front)++; 
return queue[*front - 1]; 
} 

Answer 1

the variable rear is a pointer to an int. With *rear you get the value of that int. That value is then used as index.

It's equivalent to e.g.

int index = *rear;
queue[index] = new_node;
index++;
*rear = index;

It copies the value of the variable new_node (i.e. the address of where the pointer is pointing) into queue[index]. From this point onward, both new_node and queue[index] points to the same thing.

I hope that makes it clearer what's happening.