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Harmony Methods and Pointers

A method m with argument a is invoked in its most basic form as follows (assigning the result to r).

r = m a
That's right, no parentheses are required. In fact, if you invoke m(a), the argument is (a), which is the same as a. If you invoke m(), the argument is (), which is the empty tuple. If you invoke m(a, b), the argument is (a, b), the tuple consisting of values a and b.

You may note that all this looks familiar. Indeed, the syntax is the same as that for dictionaries and lists (see Chapter 4). Dictionaries, lists, and methods all map Harmony values to Harmony values, and their syntax is indistinguishable. If f is a method, list, or dictionary, and x is some Harmony value, then f x, f(x), and f[x] are all the same expression in Harmony.

def P_enter(pm, pid):
    pm->flags[pid] = True
    pm->turn = 1 - pid
    await (not pm->flags[1 - pid]) or (pm->turn == pid)

def P_exit(pm, pid):
    pm->flags[pid] = False

def P_mutex() returns result:
    result = { .turn: choose({0, 1}), .flags: [ False, False ] }

#### The code above can go into its own Harmony module ####

sequential mutex
mutex = P_mutex()

def thread(self):
    while choose({ False, True }):
        P_enter(?mutex, self)
        cs: assert countLabel(cs) == 1
        P_exit(?mutex, self)

spawn thread(0)
spawn thread(1)
Figure 7.1 (code/PetersonMethod.hny): Peterson's Algorithm accessed through methods

Harmony does not have a return statement. Using the returns clause of def, a result variable can be declared, for example: def f() returns something. The result of the method should be assigned to variable something. If there is no returns clause, then (for backwards compatibility reasons) the method has a default result variable called result. The default value of result is None for compatibility with Python.

Harmony also does not support break or continue statements in loops. One reason for their absence is that, particularly in concurrent programming, such control flow directions are highly error-prone. It's too easy to forget to, say, release a lock when returning a value in the middle of a method---a major source of bugs in practice.

Harmony is not an object-oriented language like Python is. In Python, you can pass a reference to an object to a method, and that method can then update the object. In Harmony, it is also sometimes convenient to have a method update a shared variable specified as an argument. For this, as mentioned in Chapter 4, each shared variable has an address, itself a Harmony value. If x is a shared variable, then the expression ?x is the address of x. If a variable contains an address, we call that variable a pointer. If p is a pointer to a shared variable, then the expression !p is the value of the shared variable. In particular, !?x = x. This is similar to how C pointers work (*&x = x).

Often, pointers point to dictionaries, and so if p is such a pointer, then (!p).field would evaluate to the specified field in the dictionary. Note that the parentheses in this expression are needed, as !p.field would wrongly evaluate !(p.field). (!p).field is such a common expression that, like C, Harmony supports the shorthand p->field, which greatly improves readability.

Figure 7.1 again shows Peterson's algorithm, but this time with methods defined to enter and exit the critical section. The name mutex is often used to denote a variable or value that is used for mutual exclusion. P_mutex is a method that returns a "mutex," which, in this case, is a dictionary that contains Peterson's Algorithm's shared memory state: a turn variable and two flags. Both methods P_enter and P_exit take two arguments: a pointer to a mutex and the thread identifier (0 or 1). pm->turn is the value of the .turn key in the dictionary that pm points to.

You can put the first three methods in its own Harmony source file and include it using the Harmony import statement. This would make the code usable by multiple applications.

Finally, methods can have local variables. Method variables are either mutable (writable) or immutable (read-only). The arguments to a method and the bound variable (or variables) within a for statement are immutable; the result variable is mutable. Using the var statement, new mutable local variables can be declared. For example, var \(x = 3\) declares a new mutable local variable x. The let statement allows declaring new immutable local variables. For example: let \(x = 3\): y \(+\)\(=\) x adds 3 to the global variable y. See for more information.

import list

current = [ [1, 2, 3], [], [] ]

while current[2] != [1, 2, 3]:
    let moves = { (s, d) for s in {0..2} for d in {0..2}
        where current[s] != []
        where (current[d] == []) or (current[s][0] < current[d][0]) }
    let (src,dst) = choose moves:
        print str(src) + " -> " + str(dst)
        current[dst] = [list.head(current[src]),] + current[dst]
        current[src] = list.tail(current[src])

assert False
Figure 7.2 (code/hanoi.hny): Towers of Hanoi

As an example of using import and let, Figure 7.2 solves the Towers of Hanoi problem. If you are not familiar with this problem: there are three towers with disks of varying sizes. In the initial configuration, the first tower has three disks (of sizes 1, 2, and 3), with the largest disk at the bottom, while the other two towers are empty. You are allowed to move a top disk from one tower to another, but you are not allowed to stack a larger disk on a smaller one. The objective is to move the disks from the first tower to the third one.

The program uses the list module documented in . It has methods to extract the head (first element) and the tail (remaining elements) of a list. (The code is simple and available in modules/list.hny.) The program tries valid moves at random until it finds a solution. Curiously, the program then asserts False. This is to cause the model checker to stop and output the trace. If you look in the output column of the trace, you will find the minimal number of moves necessary to solve the problem.

It is even cooler to remove that assertion and let Harmony generate all possible solutions to the problem like so:

$ harmony -o hanoi.png code/hanoi.hny

The resulting hanoi.png file contains a DFA describing the possible solutions. It is a little too big to include here, but well worth looking at.

const FIFO = False

def CLOCK(n) returns result:
    result = { .entries: [None,] * n, .recent: {}, .hand: 0, .misses: 0 }

def ref(clock, x):
    if x not in clock->entries:
        while clock->entries[clock->hand] in clock->recent:
            clock->recent -= {clock->entries[clock->hand]}
            clock->hand = (clock->hand + 1) % len(clock->entries)
        clock->entries[clock->hand] = x
        clock->hand = (clock->hand + 1) % len(clock->entries)
        clock->misses += 1
    if not FIFO:
        clock->recent |= {x}

clock3, clock4, refs = CLOCK(3), CLOCK(4), []

const VALUES = { 1..5 }

var last = {}
for i in {1..100}:
    let x = i if i < 5 else choose(VALUES - last):
        refs = refs + [x,]
        ref(?clock3, x); ref(?clock4, x)
        assert(clock4.misses <= clock3.misses)
        last = {x}
Figure 7.3 (code/clock.hny): Harmony program that finds page replacement anomalies

If you are ready to learn about how locks are implemented in practice, you can now skip the rest of this chapter. But if you would like to see a cool example of using the concepts introduced in this chapter, hang on for a sequential Harmony program that finds anomalies in page replacement algorithms. In 1969, Bélády et al. published a paper that showed that making a cache larger does not necessarily lead to a higher hit ratio. He showed this for a scenario using a FIFO replacement policy when the cache is full. The program in Figure 7.3 will find exactly the same scenario if you define FIFO to be True. Moreover, if you define FIFO to be False, it will find a scenario for the CLOCK replacement policy, often used in modern operating systems.

In this program, CLOCK maintains the state of a cache (in practice, typically pages in memory). The set recent maintains whether an access to the cache for a particular reference was recent or not. (It is not used if FIFO is True.) The integer misses maintains the number of cache misses. Method ref(ck, x) is invoked when x is referenced and checked against the cache ck.

The program declares two caches: one with 3 entries (clock3) and one with 4 entries (clock4). The interesting part is in the last block of code. It runs every sequence of references of up to 100 entries, using references in the range 1 through 5. Note that all the constants chosen in this program (3, 4, 5, 100) are the result of some experimentation---you can try other ones. To reduce the search space, the first four references are pinned to 1, 2, 3, and 4. Further reducing the search space, the program never repeats the same reference twice in a row (using the local variable last).

The two things to note here is the use of the choose expression and the assert statement. Using choose, we are able to express searching through all possible strings of references without a complicated nested iteration. Using assert, we are able to express the anomaly we are looking for.

In case you want to check if you get the right results. For FIFO, the program finds the same anomaly that Bélády et al. found: 1 2 3 4 1 2 5 1 2 3 4 5. For the CLOCK algorithm the program actually finds a shorter reference string: 1 2 3 4 2 1 2 5 1 2.


7.1 (This is just for fun or exercise as it is not a concurrent or distributed problem.) Implement a Harmony program that finds solutions to the "cabbage, goat, and wolf" problem. In this problem, a person accompanied by these three items has to cross a stream in a small boat, but can only take one item at a time. So, the person has to cross back and forth several times, leaving two items on one or the other shore by themselves. Unfortunately, if left to themselves, the goat would eat the cabbage and the wolf would eat the goat. What crossings does the person need to make in order not to lose any items?