Question

Here is some code from a lock-free queue using compareAndSet (in Java):

public void enq(T value) {
    Node newNode = new Node(value);
    while(true) {
        Node last = tail.get();
        Node next = last.next.get();

        if(last != tail.get())
            continue;   //???       

        if (next != null) { //improve tail
            tail.compareAndSet(last, next);
            continue;
        }

        if (last.next.compareAndSet(null, newNode)) {   //update last node
            tail.compareAndSet(last, newNode);  //update tail
            return;
        }
    }
}

public T deq() throws EmptyException {
    while(true) {
        Node first = head.get();
        Node last = tail.get();
        Node next = first.next.get();

        if(first != head.get())
            continue;   //???

        if(first == last) {
            if (next == null)
                throw new EmptyException();

            tail.compareAndSet(last, next);
            continue;
        }

        T value = next.value;
        if (head.compareAnsdSet(first, next)) {
            return value;
        }
    }
}

(head and tail are the members of the queue)

In both the deq and enq function, the first check seems unnecessary to me. (The ones commented with “???”)
I suspect it’s there just for some kind of optimization.

Am I missing something here? Do these checks affect the correctness of the code?

(The code is taken from the “Art of Multi-processor Programming”, though I did refactor the code style to have less nested ifs and elses, while keeping the code equivalent)

Answer 1

Yeah, in Java, given that it has garbage collection, those ifs have only real value as optimization, and it’s kinda a big one: CAS is incredibly expensive compared to just a read from memory, so making sure the value hasn’t changed in the meantime, and thus decreasing the chance of failing on the subsequent CAS, helps reduce the number of CAS-retries, which helps performance.

You can also move the first == last && tail-update check to inside the head.CAS, as a further optimization: the tail can lag behind only if the head was updated, so checking that only if the CAS succeeded makes sense. You can also move tail.get there then, as you don’t need it anywhere else. Example code below. Hope this helps!

public T deq() throws EmptyException {
while(true) {
    Node first = head.get();
    Node next = first.next.get();

    if (first != head.get())
        continue;

    if (next == null) {
        throw new EmptyException();
    }

    T value = next.value;

    if (head.compareAndSet(first, next)) {
        Node last = tail.get();

        if (first == last) {
            tail.compareAndSet(last, next);
        }

        return value;
    }
}

}