在LinkedList中出队

Question

I'm trying to get my dequeue method working on my implementation of a LinkedList ADT. However, it is removing from the beginning of the queue instead of the end. Any help with this? I'm new to C, and am trying to port a java exercise over to C.. It's supposed to remove the last node of the list.

Here's my dequeue method:

static void linkedQueueDequeue(Queue* q) {
    LinkedQueue *lq = ((LinkedQueue*)q->privateData);
    Node* temp = lq->head->next;
    lq->head->data = lq->head->next->data;
    lq->head->next = temp->next;
    free(temp);
    lq->size--;


}

Here's the output when trying to dequeue last node:

=====================
|Testing LinkedQueue|
=====================
adding 1 to first node
adding 2 to second node
adding 3 to third node
adding 4 to fourth node
[1,2,3,4]
dequeue last node
should print [1,2,3]
[2,3,4]
return first node
peek: 2
what's the size?
size: 3

Answer 1

As you saw already, the code in linkedQueueDequeue pops the first entry as if you wanted a stack (LIFO), you can iterate your temp to the end of the list, then remove it's temp->next :

static void linkedQueueDequeue(Queue* q) {
    LinkedQueue *lq = ((LinkedQueue*)q->privateData);
    Node* temp = lq->head->next;
    while(temp->next) temp = temp->next;
    free(temp->next);
    temp->next = 0;
    lq->size--;
}

Also note, that there ist something slightly odd about your Queue/LinkedQueue implementation considering the conversion (LinkedQueue*)q in line 2. Are you sure you need that cast? I cannot really tell because you did not give us the definitions of Queue and LinkedQueue. Is there maybe also a ->tail in LinkedQueue ? If so, then you dont need the iteration and can instead use ->tail to position temp (and of course: you have to update ->tail to the new end).

Answer 2

Your output appears to be the correct behavior for a queue/FIFO in that the first item to be removed from the list is the first item that was added to the list. Are you trying instead to create a stack?