Lecture 21: A Bounded Queue


Proof of Concept for final project due this Friday.


  1. Testing Previous Queues
  2. A Bounded Queue
  3. Stacks


  1. Unbounded Queue with Locks
    • lock enq and deq methods
  2. Lock-free Unbounded Queue
    • use AtomicReference for head, tail, next
    • use compareAndSet to validate & update atomically
    • update head/tail proactively
      • guard against incomplete deq/enq

Lock-free Dequeue Method

    public T deq() throws EmptyException {
	while (true) {
	    Node first = head.get();
	    Node last = tail.get();
	    Node next = first.next.get();
	    if (first == head.get()) {
		if (first == last) {
		    if (next == null) {
			throw new EmptyException();
		    tail.compareAndSet(last, next);
		} else {
		    T value = next.value;
		    if (head.compareAndSet(first, next))
			return value;

Testing our Queues

Bounded Queues

  • Both previous queues were unbounded
  • Sometimes we want bounded queues:
    • have limited space
    • want to force tighter synchronization between producers & consumers

How can we implement a bounded queue (with locks)?

One Option

Keep track of size!

  • Start with our UnboundedQueue
    • lock enq and deq methods
  • Add a final int capacity field
  • Add an AtomicInteger size field
    • increment upon enq
    • decrement upon deq
  • Make sure size <= capacity

Why should size be atomic?

Enqueue Method

  1. acquires enqLock
  2. if size is less than capacity
    • enqueue item
    • increment size
    • release lock
  3. otherwise
    • throw exception? (total method)
    • wait until size < capacity? (partial method)

Dequeue Method

  1. acquires deqLock
  2. if size is greater than 0
    • dequeue item
    • decrement size
    • release lock
  3. otherwise
    • throw exception? (total method)
    • wait until size > 0? (partial method)

An Unexceptional Queue

Suppose we don’t want to throw exceptions

  • Full/empty queue operations are expected, not exceptional
  • Queue should handle these cases by having threads wait

Question: How might we implement this behavior?

Enqueue with Waiting

    public void enq (T value) {
	try {
	    Node nd = new Node(value);
            while (size.get() == capacity) { }; // wait until not full
	    tail.next = nd;
	    tail = nd;
	} finally {

A Problem?

This is wasteful!

while (size.get() == capacity) { }; // wait until not full

The thread:

  1. Acquires lock
  2. Fails to enqueue while holding lock
  3. Uses resources to repeatedly check condition size.get() == capacity

What if it takes a while until the queue is not full?

A More Prudent Way

The following would be better:

  1. enqueuer sees that size.get() == capacity
  2. enqueuer temporarily gives up lock
  3. enqueuer passively waits for the condition that size.get() < capacity
    • not constantly checking
  4. dequeuer sees that enqueuers are waiting
  5. after dequeue, dequeuer notifies waiting enqueuers
  6. enqueuer acquires lock, enqueues

A Waiting Room Analogy

Queue Full

Enqueuer Arrives, Acquires Lock

Enqueuer Sees Full, Waits

Enqueuer Arrives, Acquires Lock

Enqueuer Sees Full, Waits

Dequeuer Arrives, Sees Full, Dequeues

Dequeuer Announces No Longer Full

Enqueuers Leaving Waiting Room

Dequeuer Releases Lock

Enqueuer Locks; Enqueue Ensues

The Lock and Condition Interfaces

The Lock interface defines a curious method:

  • Condition newCondition() returns a new Condition instance that is bound to this Lock instance

The Condition interface

  • void await() causes the current thread to wait until it is signalled or interrupted
  • void signal() wakes up one waiting thread
  • void signalAll() wakes up all waiting threads

Improving our Queue

Define condition for enqLock

  • notFullCondition

When enqueuing to a full queue

  • thread signals that it is waiting
    • need a flag (volatile boolean) for this
  • thread calls notFullCondition.await()
    • waits until notFullCondition is satisfied

When Dequeueing

When dequeueing

  • thread checks if thread is waiting
    • checks flag for this
  • if so, after dequeue, dequeuer calls notFullCondition.signalAll()
    • signals that queue is no longer full

(Similar: notEmptyCondition for deq method)

Making a BoundedQueue

public class BoundedQueue<T> implements SimpleQueue<T> {
    ReentrantLock enqLock, deqLock;
    Condition notEmptyCondition, notFullCondition;
    AtomicInteger size;
    volatile Node head, tail;
    final int capacity;

    public BoundedQueue(int capacity) {
	this.capacity = capacity;
	this.head = new Node(null);
	this.tail = this.head;
	this.size = new AtomicInteger(0);
	this.enqLock = new ReentrantLock();
	this.notFullCondition = this.enqLock.newCondition();
	this.deqLock = new ReentrantLock();
	this.notEmptyCondition = this.deqLock.newCondition();

    public void enq(T item) { ... }

    public T deq() { ... }

    class Node { ... }


    public void enq(T item) {
	boolean mustWakeDequeuers = false;
	Node nd = new Node(item);
	try {
	    while (size.get() == capacity) {
		try {
		    // System.out.println("Queue full!");
		} catch (InterruptedException e) {
		    // do nothing
	    tail.next = nd;
	    tail = nd;
	    if (size.getAndIncrement() == 0) {
		mustWakeDequeuers = true;
	} finally {

	if (mustWakeDequeuers) {
	    try {
	    } finally {


    public T deq() {
	T item;
	boolean mustWakeEnqueuers = false;
	try {

	    while (head.next == null) {
		try {
		    // System.out.println("Queue empty!");
		} catch(InterruptedException e) {
		    //do nothing
	    item = head.next.item;
	    head = head.next;

	    if (size.getAndDecrement() == capacity) {
		mustWakeEnqueuers = true;
	} finally {

	if (mustWakeEnqueuers) {
	    try {
	    } finally {

	return item;

Testing the Queue

For Your Consideration

What if we want to make a lock-free bounded queue?

  • Can we just add size and capacity fields to our LockFreeQueue?


Recall the Stack

Basic operations

  • void push(T item) add a new item to the top of the stack
  • T pop() remove top item from the stack and return it
    • throw EmptyException if stack was empty

Linked List Implementation

push() Step 1: Create Node

push() Step 2: Set next

push() Step 3: Set head

push() Complete


pop() Step 1: Store value

pop() Step 2: Update head

pop() Step 3: Return value

Concurrent Stack

With locks:

  • Since all operations modify head, coarse locking is natural choice

Without locks?

A Lock-free Stack

Use linked-list implementation

  • Logic is simpler than queues’ because all operations affect same node
  • Idea:
    • store top as an AtomicReference<Node>
    • use compareAndSet to modify top
      • success, or retry
  • Unlike queue:
    • item on top of stack precisely when top points to item’s Node

Implementing the Lock-Free Stack

public class LockFreeStack<T> implements SimpleStack<T> {
    AtomicReference<Node> top = new AtomicReference<Node>(null);

    public void push(T item) {...}

    public T pop() throws EmptyException {...}
    class Node {
	public T value;
	public AtomicReference<Node> next;
	public Node(T value) {
	    this.value = value;
	    this.next  = new AtomicReference<Node>(null);


Implementing push

    public void push(T item) {
	Node nd = new Node(item);
	Node oldTop = top.get();
	while (!top.compareAndSet(oldTop, nd)) {
	    oldTop = top.get();

Implementing pop

    public T pop() throws EmptyException {
	while (true) {
	    Node oldTop = top.get();
	    if (oldTop == null) {
		throw new EmptyException();

	    Node newTop = oldTop.next.get();

	    if (top.compareAndSet(oldTop, newTop)) {
		return oldTop.value;

Sequential Bottleneck

Modifying top

  • No matter how many threads, push/pop rate limited by top.compareAndSet(...)
  • This seems inherent to any stack…

… or is it?


Consider several concurrent accesses to a stack:

  • T1 calls stk.push(item1)
  • T2 calls stk.push(item2)
  • T3 calls stk.pop()
  • T4 calls stk.push(item4)
  • T5 calls stk.pop()
  • T6 calls stk.pop()

Trick Question. What is the state of stk after these calls?

What do we need the stack for?

Match and exchange!

Cut out the middleperson!

A Different Strategy

  1. Attempt to push/pop to stack
    • if success, good job
  2. If attempt fails, try to find a partner
    • if push, try to find a pop and give them your value
    • if pop, try to find a push and take their value

Next time: implement an exchange object to facilitate this