Lecture #14, Tuesday, October 24th
==================================
 Implementing `definetype` & `cases` in Schlac
 Recursive Environments
 Recursion: Racket's `letrec`
 Implementing Recursion using `letrec`
 Implementing `rec` Using Cyclic Structures
 Boxes and Mutation

# Implementing `definetype` & `cases` in Schlac
Another interesting way to implement lists follows the pattern matching
approach, where both pairs and the null value are represented by a
function that serves as a kind of a `match` dispatcher. This function
takes in two inputs  if it is the representation of null then it will
return the first input, and if it is a pair, then it will apply the
second input on the two parts of the pair. This is implemented as
follows (with type comments to make it clear):
;; null : List
(define null
(lambda (n p)
n))
;; cons : A List > List
(define cons
(lambda (x y)
(lambda (n p)
(p x y))))
This might seem awkward, but it follows the intended use of pairs and
null as a matchlike construct. Here is an example, with the equivalent
Racket code on the side:
;; Sums up a list of numbers
(define/rec (sum l)
(l ; (match l
0 ; ['() 0]
(lambda (x xs) ; [(cons x xs)
(+ x (sum xs))))) ; (+ x (sum xs))])
In fact, it's easy to implement our selectors and predicate using this:
(define null? (lambda (l) (l #t (lambda (x xs) #f))))
(define car (lambda (l) (l #f (lambda (x xs) x))))
(define cdr (lambda (l) (l #f (lambda (x xs) xs))))
;; in the above `#f' is really any value, since it
;; should be an error alternatively:
(define car (lambda (l)
(l ((lambda (x) (x x)) (lambda (x) (x x))) ; "error"
(lambda (x y) x))))
The same approach can be used to define any kind of new data type in a
way that looks like our own `definetype` definitions. For example,
consider a muchsimplified definition of the AE type we've seen early in
the semester, and a matching `eval` definition as an example for using
`cases`:
(definetype AE
[Num Number]
[Add AE AE])
(: eval : AE > Number)
(define (eval expr)
(cases expr
[(Num n) n]
[(Add l r) (+ (eval l) (eval r))]))
We can follow the above approach now to write Schlac code that more than
being equivalent, is also very similar in nature. Note that the type
definition is replaced by two definitions for the two constructors:
(define Num (lambda (n) (lambda (num add) (num n ))))
(define Add (lambda (l r) (lambda (num add) (add l r))))
(define/rec eval
(lambda (expr) ; `expr` is always a (lambda (num add) ...), and it
; expects a unary `num` argument and a binary `add`
(expr (lambda (n) n)
(lambda (l r) (+ (eval l) (eval r))))))
(test (>nat (eval (Add (Num 1) (Num 2)))) => '3)
We can even take this further: the translations from `definetype` and
`cases` are mechanical enough that we could implement them almost
exactly via rewrites (there are a subtle change in that we're list field
names rather than types):
(rewrite (definetype ignoredType [Variant arg ...] ...)
=> (define Variant
(lambda (arg ...)
(lambda (Variant ...) (Variant arg ...))))
...)
(rewrite (cases value [(ignoredVariant arg ...) result] ...)
=> (value (lambda (arg ...) result) ...))
And using that, an evluator is simple:
(definetype AE [Num n] [Add l r] [Sub l r] [Mul l r])
(define/rec eval
(lambda (expr)
(cases expr
[(Num n) n]
[(Add l r) (+ (eval l) (eval r))]
[(Sub l r) ( (eval l) (eval r))]
[(Mul l r) (* (eval l) (eval r))])))
(test (>nat (eval (Mul (Add (Num 1) (Num 2))
(Sub (Num 4) (Num 2)))))
=> '6)

# Recursive Environments
> [PLAI §11.5]
What we really need for recursion, is a special kind of an environment,
one that can refer to itself. So instead of doing (note: `call`s removed
for readability):
{with {fact {fun {n}
{if {zero? n} 1 {* n {fact { n 1}}}}}}
{fact 5}}
which does not work for the usual reasons, we want to use some
{rec {fact {fun {n}
{if {zero? n} 1 {* n {fact { n 1}}}}}}
{fact 5}}
that will do the necessary magic.
One way to achieve this is using the Y combinator as we have seen  a
kind of a "constructor" for recursive functions. We can do that in a
similar way to the `rewrite` rule that we have seen in Schlac 
translate the above expression to:
{with {fact {makerec
{fun {fact}
{fun {n}
{if {zero? n} 1 {* n {fact { n 1}}}}}}}}
{fact 5}}
or even:
{with {fact {{fun {f} {{fun {x} {f {x x}}} {fun {x} {f {x x}}}}}
{fun {fact}
{fun {n}
{if {zero? n} 1 {* n {fact { n 1}}}}}}}}
{fact 5}}
Now, we will see how it can be used in *our* code to implement a
recursive environment.
If we look at what `with` does in
{with {fact {fun {n}
{if {zero? n} 1 {* n {call fact { n 1}}}}}}
{call fact 5}}
then we can say that to evaluate this expression, we evaluate the body
expression in an extended environment that contains `fact`, even if a
bogus one that is good for `0` only  the new environment is created
with something like this:
extend("fact", makefactclosure(), env)
so we can take this whole thing as an operation over `env`
addfact(env) := extend("fact", makefactclosure(), env)
This gives us the firstlevel fact. But `fact` itself is still undefined
in `env`, so it cannot call itself. We can try this:
addfact(addfact(env))
but that still doesn't work, and it will never work no matter how far we
go:
addfact(addfact(addfact(addfact(addfact(...env...)))))
What we really want is infinity: a place where addfact works and the
result is the same as what we've started with  we want to create a
"magical" environment that makes this possible:
let magicenv = ???
such that:
addfact(magicenv) = magicenv
which basically gives us the illusion of being at the infinity point.
This magicenv thing is exactly the *fixedpoint* of the `addfact`
operation. We can use:
magicenv = rec(addfact)
and following the main property of the Y combinator, we know that:
magicenv = rec(addfact) ; def. of magicenv
= addfact(rec(addfact)) ; Y(f) = f(Y(f))
= addfact(magicenv) ; def. of magicenv
What does all this mean? It means that if we have a fixedpoint operator
at the level of the implementation of our environments, then we can use
it to implement a recursive binder. In our case, this means that a
fixpoint in Racket can be used to implement a recursive language. But we
have that  Racket does have recursive functions, so we should be able
to use that to implement our recursive binder.
There are two ways that make it possible to write recursive functions in
Racket. One is to define a function, and use its name to do a recursive
call  using the Racket formal rules, we can see that we said that we
mark that we now *know* that a variable is bound to a value. This is
essentially a sideeffect  we modify what we know, which corresponds
to modifying the global environment. The second way is a new form:
`letrec`. This form is similar to `let`, except that the scope that is
established includes the named expressions  it is exactly what we
want `rec` to do. A third way is using recursive local definitions, but
that is equivalent to using `letrec`, more on this soon.

# Recursion: Racket's `letrec`
So we want to add recursion to our language, practically. We already
know that Racket makes it possible to write recursive functions, which
is possible because of the way it implements its "global environment":
our evaluator can only *extend* an environment, while Racket *modifies*
its global environment. This means that whenever a function is defined
in the global environment, the resulting closure will have it as its
environment "pointer", but the global environment was not extended 
it stays the same, and was just modified with one additional binding.
But Racket has another, a bit more organized way of using recursion:
there is a special localbinding construct that is similar to `let`, but
allows a function to refer to itself. It is called `letrec`:
(letrec ([fact (lambda (n)
(if (zero? n)
1
(* n (fact ( n 1)))))])
(fact 5))
Some people may remember that there was a third way for creating
recursive functions: using local definition in function bodies. For
example, we have seen things like:
(define (length list)
(define (helper list len)
(if (null? list)
len
(helper (rest list) (+ len 1))))
(helper list 0))
This looks like the same kind of environment magic that happens with a
global `define`  but actually, Racket defines the meaning of internal
definitions using `letrec`  so the above code is exactly the same as:
(define (length list)
(letrec ([helper (lambda (list len)
(if (null? list)
len
(helper (rest list) (+ len 1))))])
(helper list 0)))
The scoping rules for a `letrec` is that the scope of the bound name
covers both the body *and* the named expression. Furthermore, multiple
names can be bound to multiple expressions, and the scope of each name
covers all named expression as well as the body. This makes it easy to
define mutually recursive functions, such as:
(letrec ([even? (lambda (n) (if (zero? n) #t (odd? ( n 1))))]
[odd? (lambda (n) (if (zero? n) #f (even? ( n 1))))])
(even? 99))
But it is not a required functionality  it could be done with a
single recursive binding that contains several functions:
(letrec ([even+odd
(list (lambda (n)
(if (zero? n) #t ((second even+odd) ( n 1))))
(lambda (n)
(if (zero? n) #f ((first even+odd) ( n 1)))))])
((first even+odd) 99))
This is basically the same problem we face if we want to use the Y
combinator for mutually recursive bindings. The above solution is
inconvenient, but it can be improved using more `let`s to have easier
name access. For example:
(letrec ([even+odd
(list (lambda (n)
(let ([even? (first even+odd)]
[odd? (second even+odd)])
(if (zero? n) #t (odd? ( n 1)))))
(lambda (n)
(let ([even? (first even+odd)]
[odd? (second even+odd)])
(if (zero? n) #f (even? ( n 1))))))])
(let ([even? (first even+odd)]
[odd? (second even+odd)])
(even? 99)))

# Implementing Recursion using `letrec`
We will see how to add a similar construct to our language  for
simplicity, we will add a `rec` form that handles a single binding:
{rec {fact {fun {n}
{if {= 0 n}
1
{* n {fact { n 1}}}}}}
{fact 5}}
Using this, things can get a little tricky. What should we get if we do:
{rec {x x} x}
? Currently, it seems like there is no point in using any expression
except for a *function* expression in a `rec` expression, so we will
handle only these cases.
(BTW, under what circumstances would nonfunction values be useful in a
letrec?)

One way to achieve this is to use the same trick that we have recently
seen: instead of reimplementing language features, we can use existing
features in our own language, which hopefully has the right
functionality in a form that can be reused to in our evaluator.
Previously, we have seen a way to implement environments using Racket
closures:
;; Define a type for functional environments
(definetype ENV = Symbol > VAL)
(: EmptyEnv : > ENV)
(define (EmptyEnv)
(lambda (id) (error 'lookup "no binding for ~s" id)))
(: lookup : Symbol ENV > VAL)
(define (lookup name env)
(env name))
(: Extend : Symbol VAL ENV > ENV)
(define (Extend id val restenv)
(lambda (name)
(if (eq? name id)
val
(restenv name))))
We can use this implementation, and create circular environments using
Racket's `letrec`. The code for handling a `with` expressions is:
[(With boundid namedexpr boundbody)
(eval boundbody
(Extend boundid (eval namedexpr env) env))]
It looks like we should be able to handle `rec` in a similar way (the
AST constructor name is `WRec` ("withrec") so it doesn't collide with
TR's `Rec` constructor for recursive types):
[(WRec boundid namedexpr boundbody)
(eval boundbody
(Extend boundid (eval namedexpr env) env))]
but this won't work because the named expression is evaluated
prematurely, in the previous environment. Instead, we will move
everything that needs to be done, including evaluation, to a separate
`extendrec` function:
[(WRec boundid namedexpr boundbody)
(eval boundbody
(extendrec boundid namedexpr env))]
Now, the `extendrec` function needs to provide the new, "magically
circular" environment. Following what we know about the arguments to
`extendrec`, and the fact that it returns a new environment (= a lookup
function), we can sketch a rough definition:
(: extendrec : Symbol FLANG ENV > ENV) ; FLANG, not VAL!
;; extend an environment with a new binding that is the result of
;; evaluating an expression in the same environment as the extended
;; result
(define (extendrec id expr restenv)
(lambda (name)
(if (eq? name id)
... something that uses expr to get a value ...
(restenv name))))
What should the missing expression be? It can simply evaluate the object
given itself:
(define (extendrec id expr restenv)
(lambda (name)
(if (eq? name id)
(eval expr ...this environment...)
(restenv name))))
But how do we get *this environment*, before it is defined? Well, the
environment is itself a Racket *function*, so we can use Racket's
`letrec` to make the function refer to itself recursively:
(define (extendrec id expr restenv)
(letrec ([recenv (lambda (name)
(if (eq? name id)
(eval expr recenv)
(restenv name)))])
recenv))
It's a little more convenient to use an internal definition, and add a
type for clarity:
(define (extendrec id expr restenv)
(: recenv : Symbol > VAL)
(define (recenv name)
(if (eq? name id)
(eval expr recenv)
(restenv name)))
recenv)
This works, but there are several problems:
1. First, we no longer do a simple lookup in the new environment.
Instead, we evaluate the expression on *every* such lookup. This
seems like a technical point, because we do not have sideeffects in
our language (also because we said that we want to handle only
function expressions). Still, it wastes space since each evaluation
will allocate a new closure.
2. Second, a related problem  what happens if we try to run this:
{rec {x x} x}
? Well, we do that stuff to extend the current environment, then
evaluate the body in the new environment, this body is a single
variable reference:
(eval (Id 'x) thenewenv)
so we look up the value:
(lookup 'x thenewenv)
which is:
(thenewenv 'x)
which goes into the function which implements this environment, there
we see that `name` is the same as `name1`, so we return:
(eval expr recenv)
but the `expr` here is the original namedexpression which is itself
`(Id 'x)`, and we're in an infinite loop.
We can try to get over these problems using another binding. Racket
allows several bindings in a single `letrec` expression or multiple
internal function definitions, so we change `extendrec` to use the
newlycreated environment:
(define (extendrec id expr restenv)
(: recenv : Symbol > VAL)
(define (recenv name)
(if (eq? name id)
val
(restenv name)))
(: val : VAL)
(define val (eval expr recenv))
recenv)
This runs into an interesting type error, which complains about possibly
getting some `Undefined` value. It does work if we switch to the untyped
language for now (using `#lang pl untyped`)  and it seems to run fine
too. But it raises more questions, beginning with: what is the meaning
of:
(letrec ([x ...]
[y ...x...])
...)
or equivalently, an internal block of
(define x ...)
(define y ...x...)
? Well, DrRacket seems to do the "right thing" in this case, but what
about:
(letrec ([y ...x...]
[x ...])
...)
? As a hint, see what happens when we now try to evaluate the
problematic
{rec {x x} x}
expression, and compare that with the result that you'd get from Racket.
This also clarifies the type error that we received.
It should be clear now why we want to restrict usage to just binding
recursive functions. There are no problems with such definitions because
when we evaluate a `fun` expression, there is no evaluation of the body,
which is the only place where there are potential references to the same
function that is defined  a function's body is *delayed*, and
executed only when the function is applied later.
But the biggest question that is still open: we just implemented a
circular environment using Racket's own circular environment
implementation, and that does not explain how they are actually
implemented. The cycle of pointers that we've implemented depends on the
cycle of pointers that Racket uses, and that is a black box we want to
open up.
For reference, the complete code is below.
#lang pl
#
The grammar:
::=
 { + }
 {  }
 { * }
 { / }
 { with { } }
 { rec { } }

 { fun { } }
 { call }
Evaluation rules:
eval(N,env) = N
eval({+ E1 E2},env) = eval(E1,env) + eval(E2,env)
eval({ E1 E2},env) = eval(E1,env)  eval(E2,env)
eval({* E1 E2},env) = eval(E1,env) * eval(E2,env)
eval({/ E1 E2},env) = eval(E1,env) / eval(E2,env)
eval(x,env) = lookup(x,env)
eval({with {x E1} E2},env) = eval(E2,extend(x,eval(E1,env),env))
eval({rec {x E1} E2},env) = ???
eval({fun {x} E},env) = <{fun {x} E}, env>
eval({call E1 E2},env1)
= eval(Ef,extend(x,eval(E2,env1),env2))
if eval(E1,env1) = <{fun {x} Ef}, env2>
= error! otherwise
#
(definetype FLANG
[Num Number]
[Add FLANG FLANG]
[Sub FLANG FLANG]
[Mul FLANG FLANG]
[Div FLANG FLANG]
[Id Symbol]
[With Symbol FLANG FLANG]
[WRec Symbol FLANG FLANG]
[Fun Symbol FLANG]
[Call FLANG FLANG])
(: parsesexpr : Sexpr > FLANG)
;; parses sexpressions into FLANGs
(define (parsesexpr sexpr)
(match sexpr
[(number: n) (Num n)]
[(symbol: name) (Id name)]
[(cons 'with more)
(match sexpr
[(list 'with (list (symbol: name) named) body)
(With name (parsesexpr named) (parsesexpr body))]
[else (error 'parsesexpr "bad `with' syntax in ~s" sexpr)])]
[(cons 'rec more)
(match sexpr
[(list 'rec (list (symbol: name) named) body)
(WRec name (parsesexpr named) (parsesexpr body))]
[else (error 'parsesexpr "bad `rec' syntax in ~s" sexpr)])]
[(cons 'fun more)
(match sexpr
[(list 'fun (list (symbol: name)) body)
(Fun name (parsesexpr body))]
[else (error 'parsesexpr "bad `fun' syntax in ~s" sexpr)])]
[(list '+ lhs rhs) (Add (parsesexpr lhs) (parsesexpr rhs))]
[(list ' lhs rhs) (Sub (parsesexpr lhs) (parsesexpr rhs))]
[(list '* lhs rhs) (Mul (parsesexpr lhs) (parsesexpr rhs))]
[(list '/ lhs rhs) (Div (parsesexpr lhs) (parsesexpr rhs))]
[(list 'call fun arg)
(Call (parsesexpr fun) (parsesexpr arg))]
[else (error 'parsesexpr "bad syntax in ~s" sexpr)]))
(: parse : String > FLANG)
;; parses a string containing a FLANG expression to a FLANG AST
(define (parse str)
(parsesexpr (string>sexpr str)))
;; Types for environments, values, and a lookup function
(definetype VAL
[NumV Number]
[FunV Symbol FLANG ENV])
;; Define a type for functional environments
(definetype ENV = Symbol > VAL)
(: EmptyEnv : > ENV)
(define (EmptyEnv)
(lambda (id) (error 'lookup "no binding for ~s" id)))
(: lookup : Symbol ENV > VAL)
;; lookup a symbol in an environment, return its value or throw an
;; error if it isn't bound
(define (lookup name env)
(env name))
(: Extend : Symbol VAL ENV > ENV)
;; extend a given environment cache with a new binding
(define (Extend id val restenv)
(lambda (name)
(if (eq? name id)
val
(restenv name))))
(: extendrec : Symbol FLANG ENV > ENV)
;; extend an environment with a new binding that is the result of
;; evaluating an expression in the same environment as the extended
;; result
(define (extendrec id expr restenv)
(: recenv : Symbol > VAL)
(define (recenv name)
(if (eq? name id)
val
(restenv name)))
(: val : VAL)
(define val (eval expr recenv))
recenv)
(: NumV>number : VAL > Number)
;; convert a FLANG runtime numeric value to a Racket one
(define (NumV>number val)
(cases val
[(NumV n) n]
[else (error 'arithop "expected a number, got: ~s" val)]))
(: arithop : (Number Number > Number) VAL VAL > VAL)
;; gets a Racket numeric binary operator, and uses it within a NumV
;; wrapper
(define (arithop op val1 val2)
(NumV (op (NumV>number val1) (NumV>number val2))))
(: eval : FLANG ENV > VAL)
;; evaluates FLANG expressions by reducing them to values
(define (eval expr env)
(cases expr
[(Num n) (NumV n)]
[(Add l r) (arithop + (eval l env) (eval r env))]
[(Sub l r) (arithop  (eval l env) (eval r env))]
[(Mul l r) (arithop * (eval l env) (eval r env))]
[(Div l r) (arithop / (eval l env) (eval r env))]
[(With boundid namedexpr boundbody)
(eval boundbody
(Extend boundid (eval namedexpr env) env))]
[(WRec boundid namedexpr boundbody)
(eval boundbody
(extendrec boundid namedexpr env))]
[(Id name) (lookup name env)]
[(Fun boundid boundbody)
(FunV boundid boundbody env)]
[(Call funexpr argexpr)
(let ([fval (eval funexpr env)])
(cases fval
[(FunV boundid boundbody fenv)
(eval boundbody
(Extend boundid (eval argexpr env) fenv))]
[else (error 'eval "`call' expects a function, got: ~s"
fval)]))]))
(: run : String > Number)
;; evaluate a FLANG program contained in a string
(define (run str)
(let ([result (eval (parse str) (EmptyEnv))])
(cases result
[(NumV n) n]
[else (error 'run "evaluation returned a nonnumber: ~s"
result)])))
;; tests
(test (run "{call {fun {x} {+ x 1}} 4}")
=> 5)
(test (run "{with {add3 {fun {x} {+ x 3}}}
{call add3 1}}")
=> 4)
(test (run "{with {add3 {fun {x} {+ x 3}}}
{with {add1 {fun {x} {+ x 1}}}
{with {x 3}
{call add1 {call add3 x}}}}}")
=> 7)
(test (run "{with {identity {fun {x} x}}
{with {foo {fun {x} {+ x 1}}}
{call {call identity foo} 123}}}")
=> 124)
(test (run "{with {x 3}
{with {f {fun {y} {+ x y}}}
{with {x 5}
{call f 4}}}}")
=> 7)
(test (run "{call {with {x 3}
{fun {y} {+ x y}}}
4}")
=> 7)
(test (run "{with {f {with {x 3} {fun {y} {+ x y}}}}
{with {x 100}
{call f 4}}}")
=> 7)
(test (run "{call {call {fun {x} {call x 1}}
{fun {x} {fun {y} {+ x y}}}}
123}")
=> 124)

# Implementing `rec` Using Cyclic Structures
> [PLAI §10]
Looking at the arrows in the environment diagrams, what we're really
looking for is a closure that has an environment pointer which is the
same environment in which it was defined. This will make it possible for
`fact` to be bound to a closure that can refer to *itself* since its
environment is the same one in which it is defined. However, so far we
have no tools that makes it possible to do this.
What we need is to create a "cycle of pointers", and so far we do not
have a way of achieving that: when we create a closure, we begin with an
environment which is saved in the slot's environment slot, but we want
that closure to be the value of a binding in that same environment.

# Boxes and Mutation
To actually implement a circular structure, we will now use
*sideeffects*, using a new kind of Racket value which supports
mutation: a box. A box value is built with the `box` constructor:
(define mything (box 7))
the value is retrieved with the `unbox' function,
(* 6 (unbox mything))
and finally, the value can be changed with the `setbox!` function.
(setbox! mything 17)
(* 6 (unbox mything))
An important thing to note is that `setbox!` is much like `display`
etc, it returns a value that is not printed in the Racket REPL, because
there is no point in using the result of a `setbox!`, it is called for
the sideeffect it generates. (Languages like C blur this distinction
between returning a value and a sideeffect with its assignment
statement.)
As a side note, we now have side effects of two kinds: mutation of
state, and I/O (at least the O part). (Actually, there is also infinite
looping that can be viewed as another form of a side effect.) This means
that we're now in a completely different world, and lots of new things
can make sense now. A few things that you should know about:
* We never used more than one expression in a function body because
there was no point in it, but now there is. To evaluate a sequence of
Racket expressions, you wrap them in a `begin` expression.
* In most places you don't actually need to use `begin`  these are
places that are said to have an *implicit* `begin`: the body of a
function (or any lambda expression), the body of a `let` (and
`let`relatives), the consequence positions in `cond`, `match`, and
`cases` clauses and more. One of the common places where a `begin` is
used is in an `if` expression (and some people prefer using `cond`
instead when there is more than a single expression).
* `cond` without an `else` in the end can make sense, if all you're
using it it for is sideeffects.
* `if` could get a single expression which is executed when the
condition is true (and an unspecified value is used otherwise), but
our language (as well as the default Racket language) always forbids
this  there are convenient special forms for a onesided `if`s:
`when` & `unless`, and they can have any number of expressions (they
have an implicit `begin`). They have an advantage of saying "this code
does some sideeffects here" more explicit.
* There is a function called `foreach` which is just like `map`, except
that it doesn't collect the list of results, it is used only for
performing side effects.
* Aliasing and the concept of "object equality": `equal?` vs `eq?`. For
example:
(: foo : (Boxof ...) (Boxof ...) > ...)
(define (foo a b)
(setbox! a 1)) ;*** this might change b, can check `eq?`
When any one of these things is used (in Racket or other languages), you
can tell that sideeffects are involved, because there is no point in
any of them otherwise. In addition, any name that ends with a `!`
("bang") is used to mark a function that changes state (usually a
function that only changes state).
So how do we create a cycle? Simple, boxes can have any value, and they
can be put in other values like lists, so we can do this:
#lang pl untyped
(define foo (list 1 (box 3)))
(setbox! (second foo) foo)
and we get a circular value. (Note how it is printed.) And with types:
#lang pl
(: foo : (List Number (Boxof Any)))
(define foo (list 1 (box 3)))
(setbox! (second foo) foo)