I’m reading Scala in Depth. In Section 2.3.2 on concurrency, the author gives the following example of a thread safe “index service” (removed trait definition for brevity):

class ImmutableService[Key, Value] {
var currentIndex = new ImmutableHashMap[Key,Value]
def lookUp(k : Key) : Option[Value] = currentIndex.get(k)
def insert(k : Key, v: Value) : Unit = synchronized {
currentIndex = currentIndex + ((k, v))
}
}


The author shows that this implementation, is much faster than an implementation which uses a value reference to a mutable map, and synchronises both lookup and insert operation.

When I was reading this code, I was wondering whether this implementation is indeed safe. After all, the assignment operation to the var itself is not synchronised – do we have a guarantee that it is atomic?

The Java Language Specification answered my question:

For the purposes of the Java programming language memory model, a single write to a non-volatile long or double value is treated as two separate writes: one to each 32-bit half. This can result in a situation where a thread sees the first 32 bits of a 64-bit value from one write, and the second 32 bits from another write.

Writes and reads of volatile long and double values are always atomic.

Writes to and reads of references are always atomic, regardless of whether they are implemented as 32-bit or 64-bit values.

Note that in general, updating 64-bit chunks of memory in the JVM may not be atomic, but reference updates specifically must be atomic as per the JLS.

This is a rather delicate point – in the general case, if reference assignments were not guaranteed to be atomic (as is sometimes the case in other languages), this example would not be correct since the reading threads may see a partially initialised reference.

Since this is such a delicate point, I think it’s worth mentioning explicitly in the book’s text.

## Understanding the Y Combinator

I am now reading Types and Programming Languages, and I’ve been trying to wrap my head around the Y combinator and why it works. This is how I explained it to myself:

We want to define a recursive function, which means the function should be able to refer to itself. Unfortunately, Lambda calculus does not give us this option, so we define a function of the form:

$g = \lambda func. \lambda param.\space\text{return something or} func(param')$

Where $param'$ is some reduced version of $param$. In order for this thing to be a recursion, we need to find a parameter $func$ for which:

$func = g(func)$

We notice that $func$ is a fixed point of $g$ by definition, hence the name fixed point combinators. We also notice that it must be a function of $g$ (otherwise it’s a constant and we can show trivially that it doesn’t work, duh), so we can write:

$Y(g) = func = g(func) = g(Y(g))$

Cool. Now we are looking for $term_0$ such that:

$Y = \lambda f.\underbrace{\text{ } f(term_0) \text{ }}_{term_0}$

Hmmm… Ok. Let’s try to name $term_0$ and pass it as a parameter to $f$, will that work?

$Y = \lambda f.\underbrace{(\lambda x. f(x))}_{term_0} term_0 =\lambda f.\underbrace{(\lambda x. f(x))}_{term_0}\underbrace{(\lambda x. f(x))}_{term_0}$

This sucks. After our change we no longer need to pass $term_0$ to $f$, but $term_1$ :

$Y = \lambda f.\underbrace{(\lambda x. f(term_1))(\lambda x. f(term_1))}_{term_1}$

Fortunately, can easily express $term_1$ using our bound variable $x$ as $(x x)$! So if we write:

$Y = \lambda f.(\lambda x. f(x x))(\lambda x. f(x x))$

We win!

We can see that the Y combinator as we defined it is a fixed point of the function $g$, just like we wanted.
Great success :)

I ignored reduction rules for the sake of simplicity, but it’s interesting to note that the book talks about another type of fixed point combinator, which should be used under different reduction rules:

$fix = \lambda f. (\lambda x. f (\lambda y. x x y))(\lambda x. f (\lambda y. x x y))$

I wonder how many fixed points $g$ has under each set of reduction rules.