PL: Lecture #18  Tuesday, March 22nd
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Lazy Evaluation: Some Issues

There are a few issues that we need to be aware of when we’re dealing with a lazy language. First of all, remember that our previous attempt at lazy evaluation has made

{with {x y}
  {with {y 1}
    x}}

evaluate to 1, which does not follow the rules of lexical scope. This is not a problem with lazy evaluation, but rather a problem with our naive implementation. We will shortly see a way to resolve this problem. In the meanwhile, remember that when we try the same in Lazy Racket we do get the expected error:

> (let ([x y])
    (let ([y 1])
      x))
reference to undefined identifier: y

A second issue is a subtle point that you might have noticed when we played with Lazy Racket: for some of the list values we have see a “.” printed. This is part of the usual way Racket displays an improper list — any list that does not terminate with a null value. For example, in plain Racket:

> (cons 1 2)
(1 . 2)
> (cons 1 (cons 2 (cons 3 4)))
(1 2 3 . 4)

In the dialect that we’re using in this course, this is not possible. The secret is that the cons that we use first checks that its second argument is a proper list, and it will raise an error if not. So how come Lazy Racket’s cons is not doing the same thing? The problem is that to know that something is a proper list, we will have to force it, which will make it not behave like a constructor.

As a side note, we can achieve some of this protection if we don’t insist on immediately checking the second argument completely, and instead we do the check when needed — lazily:

(define (safe-cons x l)
  (cons x (if (pair? l) l (error "poof"))))

Finally, there are two consequences of using a lazy language that make it more difficult to debug (or at lease take some time to get used to). First of all, control tends to flow in surprising ways. For example, enter the following into DrRacket, and run it in the normal language level for the course:

                (define (foo3 x)
                  (/ x "1"))
(define (foo2 x)
  (foo3 x))
                (define (foo1 x)
                  (list (foo2 x)))
(define (foo0 x)
  (first (foo1 x)))

        (+ 1 (foo0 3))

In the normal language level, we get an error, and red arrows that show us how where in the computation the error was raised. The arrows are all expected, except that foo2 is not in the path — why is that? Remember the discussion about tail-calls and how they are important in Racket since they are the only tool to generate loops? This is what we’re seeing here: foo2 calls foo3 in a tail position, so there is no need to keep the foo2 context anymore — it is simply replaced by foo3. (Incidentally, there is also no arrow that goes through foo1: Racket does some smart inlining, and it figures out that foo0+foo1 are simply returning the same value, so it skips foo1.)

Now switch to Lazy Racket and re-run — you’ll see no arrows at all. What’s the problem? The call of foo0 creates a promise that is forced in the top-level expression, that simply returns the first of the list that foo1 created — and all of that can be done without forcing the foo2 call. Going this way, the computation is finally running into an error after the calls to foo0, foo1, and foo2 are done — so we get the seemingly out-of-context error.

To follow what’s happening here, we need to follow how promise are forced: when we have code like

> (define (foo x) (/ x 0))
> (foo 9)

then the foo call is a strict point, since we need an actual value to display on the REPL. Since it’s in a strict position, we do the call, but when we’re in the function there is no need to compute the division result — so it is returned as a lazy promise value back to the toplevel. It is only then that we continue the process of getting an actual value, which leads to trying to compute the division and get the error.

Finally, there are also potential problems when you’re not careful about memory use. A common technique when using a lazy language is to generate an infinite list and pull out its Nth element. For example, to compute the Nth Fibonacci number, we’ve seen how we can do this:

(define fibs (cons 1 (cons 1 (map + fibs (rest fibs)))))
(define (fib n) (list-ref fibs n))

and we can also do this (reminder: letrec is the same as an internal definition):

(define (fib n)
  (letrec ([fibs (cons 1 (cons 1 (map + fibs (rest fibs))))])
    (list-ref fibs n))) ; tail-call => no need to keep `fibs`

but the problem here is that when list-ref is making its way down the list, it might still hold a reference to fibs, which means that as the list is forced, all intermediate values are held in memory. In the first of these two, this is guaranteed to happen since we have a binding that points at the head of the fibs list. With the second form things can be confusing: it might be that our language implementation is smart enough to see that fibs is not really needed anymore and release the offending reference. If it isn’t, then we’d have to do something like

(define (fib n)
  (list-ref
  (letrec ([fibs (cons 1 (cons 1 (map + fibs (rest fibs))))])
    fibs)
  n))

to eliminate it. But even if the implementation does know that there is no need for that reference, there are other tricky situations that are hard to avoid.

Side note: Racket didn’t use to do this optimization, but now it does, and the lazy language helped in clarifying more cases where references should be released. To see that, consider these two variants:

(define (nat1 n)
  (define nats (cons 1 (map add1 nats)))
  (if (number? (list-ref nats n))
    "a number"
    "not a number"))

;; we want to provide some information: show the first element
(define (nat2 n)
  (define nats (cons 1 (map add1 nats)))
  (if (number? (list-ref nats n))
    "a number"
    (error 'nat "the list starting with ~s is broken"
          (first nats))))

If we try to use them with a big input:

(nat1 300000) ; or with nat2

then nat1 would work fine, whereas nat2 will likely run into DrRacket’s memory limit and the computation will be terminated. The problem is that nat2 uses the nats value after the list-ref call, which will make a reference to the head of the list, preventing it from being garbage-collected while list-ref is cdr-ing down the list and making more cons cells materialize.

It’s still possible to show the extra information though – just save it:

;; we want to provide some information: show the first element
(define (nat3 n)
  (define nats (cons 1 (map add1 nats)))
  (define fst (first nats))
  (if (number? (list-ref nats n))
    "a number"
    (error 'nat "the list starting with ~s is broken" fst)))

It looks like it’s spending a redundant runtime cycle in the extra computation, but it’s a lazy language so this is not a problem.

Lazy Evaluation: Shell Examples

There is a very simple and elegant principle in shell programming — we get a single data type, a character stream, and many small functions, each doing a single simple job. With these small building blocks, we can construct more sequences that achieve more complex tasks, for example — a sorted frequency table of lines in a file:

sort foo | uniq -c | sort -nr

This is very much like a programming language — we get small blocks, and build stuff out of them. Of course there are swiss army knives like awk that try to do a whole bunch of stuff, (the same attitude that brought Perl to the world…) and even these respect the “stream” data type. For example, a simple { print $1 } statement will work over all lines, one by one, making it a program over an infinite input stream, which is what happens in reality in something like:

cat /dev/console | awk ...

But there is something else in shell programming that makes so effective: it is implementing a sort of a lazy evaluation. For example, compare this:

cat foo | awk '{ print $1+$2; }' | uniq

to:

cat foo | awk '{ print $1+$2; }' | uniq | head -5

Each element in the pipe is doing its own small job, and it is always doing just enough to feed its output. Each basic block is designed to work even on infinite inputs! (Even sort works on unlimited inputs…) (Soon we will see a stronger connection with lazy evaluation.)

Side note: (Alan Perlis) “It is better to have 100 functions operate on one data structure than 10 functions on 10 data structures”… But the uniformity comes at a price: the biggest problem shells have is in their lack of a recursive structure, contaminating the world with way too many hacked up solutions. More than that, it is extremely inefficient and usually leads to data being re-parsed over and over and over — each small Unix command needs to always output stuff that is human readable, but the next command in the pipe will need to re-parse that, eg, rereading decimal numbers. If you look at pipelines as composing functions, then a pipe of numeric commands translates to something like:

itoa(baz(atoi(itoa(bar(atoi(itoa(foo(atoi(inp)))))))))

and it is impossible to get rid of the redundant atoi(itoa(...))s.

Lazy Evaluation: Programming Examples

We already know that when we use lazy evaluation, we are guaranteed to have more robust programs. For example, a function like:

(define (my-if x y z)
  (if x y z))

is completely useless in Racket because all functions are eager, but in a lazy language, it would behave exactly like the real if. Note that we still need some primitive conditional, but this primitive can be a function (and it is, in Lazy Racket).

But we get more than that. If we have a lazy language, then computations are pushed around as if they were values (computations, because these are expressions that are yet to be evaluated). In fact, there is no distinction between computations and values, it just happens that some values contain “computational promises”, things that will do something in the future.

To see how this happens, we write a simple program to compute the (infinite) list of prime numbers using the sieve of Eratosthenes. To do this, we begin by defining the list of all natural numbers:

(define nats (cons 1 (map add1 nats)))

And now define a sift function: it receives an integer n and an infinite list of integers l, and returns a list without the numbers that can be divided by n. This is simple to write using filter:

(define (sift n l)
  (filter (lambda (x) (not (divides? n x))) l))

and it requires a definition for divides? — we use Racket’s modulo for this:

(define (divides? n m)
  (zero? (modulo m n)))

Now, a sieve is a function that consumes a list that begins with a prime number, and returns the prime numbers from this list. To do this, it returns a list that has the same first number, and for its tail it sifts out numbers that are divisible by the first from the original list’s tail, and calls itself recursively on the result:

(define (sieve l)
  (cons (first l) (sieve (sift (first l) (rest l)))))

Finally, the list of prime numbers is the result of applying sieve on the list of numbers from 2. The whole program is now:

#lang pl lazy

(define nats (cons 1 (map add1 nats)))

(define (divides? n m)
  (zero? (modulo m n)))

(define (sift n l)
  (filter (lambda (x) (not (divides? n x))) l))

(define (sieve l)
  (cons (first l) (sieve (sift (first l) (rest l)))))

(define primes (sieve (rest nats)))

To see how this runs, we trace modulo to see which tests are being used. The effect of this is that each time divides? is actually required to return a value, we will see a line with its inputs, and its output. This output looks quite tricky — things are computed only on a “need to know” basis, meaning that debugging lazy programs can be difficult, since things happen when they are needed which takes time to get used to. However, note that the program actually performs the same tests that you’d do using any eager-language implementation of the sieve of Eratosthenes, and the advantage is that we don’t need to decide in advance how many values we want to compute — all values will be computed when you want to see the corresponding result. Implementing this behavior in an eager language is more difficult than a simple program, yet we don’t need such complex code when we use lazy evaluation.

Note that if we trace divides? we see results that are some promise struct — these are unevaluated expressions, and they point at the fact that when divides? is used, it doesn’t really force its arguments — this happens later when these results are forced.

The analogy with shell programming using pipes should be clear now — for example, we have seen this:

cat foo | awk '{ print $1+$2; }' | uniq | head -5

The last head -5 means that no computation is done on parts of the original file that are not needed. It is similar to a (take 5 l) expression in Lazy Racket.