Why return the queue when using dequeue?

Question

I am having some confusion regarding queues. When I return a pointer to the start of a queue as shown in the code below, and use

Q = enQ(Q)

the function works fine. However, I don't see why the function needs to return anything, since the pointer to queue is being updated within. Why does the following not seem to work when the function is made to return void?

enQ(Q)

Code:

// Adds item to queue
struct node* enQ(struct node* Q, int n){

    struct node* last = Q;
    struct node* new = malloc(sizeof(struct node));
    new->data = n;
    new->next = NULL;

    if (!Q){
        Q = new;
    } else {
        while (last->next){
            last = last->next;
        }
        last->next = new;
    }
    return Q;
}

John Kugelman · Accepted Answer · 2018-10-18 02:02:00Z

1

The first time you call enQ() it's with a null pointer indicating that a new queue should be created. enQ() creates a new node and returns a pointer to it.

On subsequent calls you are correct that the return value isn't needed as it simply returns back the same Q that was passed in, but the first time the return value is indeed necessary. Without it calling enQ(NULL, i) won't have any way to return the new queue to the caller.

answered Oct 18, 2018 at 2:02

John Kugelman

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4 Comments

user10257574 Over a year ago

Thanks for the answer. I must have some fundamental misunderstanding about how pointers are changed in functions. In the Q = new line, is the value of Q not set to new automatically, and hence a return is unnecessary?

Sisir Over a year ago

So basically, you are doing 2 different operations in the method: queue creation and updation. While updating the queue of course you are updating the content of what Q* is pointing to, so it is okay. But when you created a new queue your local pointer is only pointing to this new queue. How will the calling function know about it, unless you return the pointer or you make the input as Q**.

user10257574 Over a year ago

Ah, so the pointer to allocated memory new is a 'local' variable, in the same way that an int would be, and therefore needs to be returned to the function?

John Kugelman Over a year ago

That is correct. If you have an assignment and dereference like *Q = ... then the caller might see the change. But a bare Q = ... only changes the local variable.

sundb · Accepted Answer · 2018-10-18 03:19:21Z

0

You don't use pass-through Node to enter enQ, you should create a Queue struct and pass in Queue*

typedef struct Queue {
    struct node* head;
} Queue;

// Adds item to queue
void enQ(struct Queue* Q, int n) {
    struct node* tail;

    struct node* new = malloc(sizeof(struct node));
    new->data = n;
    new->next = NULL;

    if (Q->head == NULL) {
        Q->head = new;
    } else {
        tail = Q->head;
        while (tail->next) {
            tail = tail->next;
        }

        tail->next = new;
    }
}

void deQ(struct Queue* Q, int n) {
    if (Q->head == NULL) {
        return;
    }

    struct node* temp = Q->head;
    Q->head = Q->head->next;
    free(temp);
}

edited Oct 18, 2018 at 3:19

answered Oct 18, 2018 at 3:13

sundb

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2 Comments

Sisir Over a year ago

What happens when Q is NULL in if (Q->head == NULL)? Don't you need to check if(Q == NULL)?

sundb Over a year ago

@Sisir No need, if you do this, there are too many things to do.

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