Types for Boxes
Obviously, Any
is not too great — it is the most generic type, so it
provides the least information. For example, notice that
returns the right list, which is equal to foo
itself — but if we try
to grab some part of the resulting list:
we get a type error, because the result of the unbox
is Any
, so
Typed Racket knows nothing about it, and won’t allow you to treat it as
a list. It is not too surprising that the type constructor that can help
in this case is Rec
which we have already seen — it allows a type
that can refer to itself:
(: foo : (Rec this (List Number (Boxof (U #f this)))))
(define foo (list 1 (box #f)))
(setbox! (second foo) foo)
Note that either foo
or the value in the box are both printed with a
Rec
type — the value in the box can’t just have a (U #f this)
type, since this
doesn’t mean anything in there, so the whole type
needs to still be present.
There is another issue to be aware of with Boxof
types. For most type
constructors (like Listof
), if T1
is a subtype of T2
, then we also
know that(Listof T1)
is a subtype of (Listof T2)
. This makes the
following code typecheck:
(: foo : (Listof Number) > Number)
(define (foo l)
(first l))
(: bar : Integer > Number)
(define (bar x)
(foo (list x)))
Since the (Listof Integer)
is a subtype of the (Listof Number)
input
for foo
, the application typechecks. But this is not the same for
the output type, for example — if we change the bar
type to:
we get a type error since Number
is not a subtype of Integer
. So
subtypes are required to “go higher” on the input side and “lower” on
the other. So, in a sense, the fact that boxes are mutable means that
their contents can be considered to be on the other side of the arrow,
which is why for such T1
subtype of T2
, it is (Boxof T2)
that is a
subtype of (Boxof T1)
, instead of the usual. For example, this doesn’t
work:
(: foo : (Boxof Number) > Number)
(define (foo b)
(unbox b))
(: bar : Integer > Number)
(define (bar x)
(: b : (Boxof Integer))
(define b (box x))
(foo b)) ;***
And you can see why this is the case — the marked line is fine given a
Number
contents, so if the type checker allows passing in a box
holding an integer, then that expression would mutate the contents and
make it an invalid value.
However, boxes are not only mutable, they hold a value that can be read
too, which means that they’re on both sides of the arrow, and this
means that (Boxof T1)
is a subtype of (Boxof T2)
if T2
is a
subtype of T1
and T1
is a subtype of T2
— in other words, this
happens only when T1
and T2
are the same type. (See below for an
extended demonstration of all of this.)
Note also that this demonstration requires that extra b
definition, if
it’s skipped:
(foo (box x)))
then this will typecheck again — Typed Racket will just consider the
context that requires a box holding a Number
, and it is still fine to
initialize such a box with an Integer
value.
As a side comment, this didn’t always work. Earlier in its existence, Typed Racket would always choose a specific type for values, which would lead to confusing errors with boxes. For example, the above would need to be written as
(define (bar x)
(foo (box (ann x : Number))))to prevent Typed Racket from inferring a specific type. This is no longer the case, but there can still be some surprises. A similar annotation was needed in the case of a list holding a selfreferential box, to avoid the initial
#f
from getting a specificbutwrong type.
Another way to see the problem these days is to enter the following expressions and see what types Typed Racket guesses for them:
> (define a 0)
> (define b (box 0))
> a
 : Integer [more precisely: Zero] ;***
0
> b
 : (Boxof Integer) ;***
'#&0Note that for
a
, the assigned type is very specific, because Typed Racket assumes that it will not change. But with a boxed value, using a type of(Boxof Zero)
would lead to a useless box, since it’ll only allow usingsetbox!
with0
, and therefore can never change. This shows that this is exactly that: a guess given the lack or explicit userspecified type, so there’s no canonical guess that can be inferred here.
Boxof
’s Lack of Subtyping
The lack of any subtype relations between (Boxof T)
and (Boxof S)
regardless of S
and T
can roughly be explained as follows.
First, a box is a container that you can pull a value out of — which makes it similar to lists. In the case of lists, we have:
then: (Listof S) subtypeof (Listof T)
This is true for all such containers that you can pull a value out of:
if you expect to pull a T
but you’re given a container of a subtype
S
, then things are still fine (you’ll get an S
which is also a T
).
Such “containers” include functions that produce a value — for
example:
then: Q > S subtypeof Q > T
However, functions also have the other side, where things are different — instead of a side of some produced value, it’s the side of the consumed value. We get the opposite rule there:
then: S > Q subtypeof T > Q
To see why this is right, use Number
and Integer
for S
and T
:
then: Number > Q subtypeof Integer > Q
so — if you expect a function that takes an integer, a valid subtype value that I can give you is a function that takes a number. In other words, every function that takes a number is also a function that takes an integer, but not the other way.
To summarize all of this, when you make the output type of a function “smaller” (more constrained), the resulting type is smaller (a subset), but on the input side things are flipped — a bigger input type means a more constrained function.
The technical names for these properties are: a “covariant” type is one that preserves the subtype relationship, and a “contravairant” type is one that reverses it. (Which is similar to how these terms are used in math.)
(Side note: this is related to the fact that in logic, P => Q
is
roughly equivalent to not(P) or Q
— the left side, P
, is inside
negation. It also explains why in ((S > T) > Q)
the S
obeys the
first rule, as if it was on the right side — because it’s negated
twice.)
Now, a (Boxof T)
is a producer of T
when you pull a value out of the
box, but it’s also a consumer of T
when you put such a value in it.
This means that — using the above analogy — the T
is on both sides
of the arrow. This means that
then: (Boxof S) subtypeof (Boxof T)
which is actually:
then: (Boxof S) isthesametypeas (Boxof T)
A different way to look at this conclusion is to consider the function
type of (A > A)
: when is it a subtype of some other (B > B)
? Only
when A
is a subtype of B
and B
is a subtype of A
, which means
that this happens only when A
and B
are the same type.
The term for this is “nonvariant” (or “invariant”):
(A > A)
is unrelated to(B > B)
regardless of howA
andB
are related. The only exception is, of course, when they are the same type. The Wikipedia entry about these puts the terms together nicely in the face of mutation:Readonly data types (sources) can be covariant; writeonly data types (sinks) can be contravariant. Mutable data types which act as both sources and sinks should be invariant.
The following piece of code makes the analogy to function types more
formally. Boxes behave as if their contents is on both sides of a
function arrow — on the right because they’re readable, and on the
left because they’re writable, which the conclusion that a (Boxof A)
type is a subtype of itself and no other (Boxof B)
.
;; a type for a "readonly" box
(definetype (Boxof/R A) = (> A))
;; Boxof/R constructor
(: box/r : (All (A) A > (Boxof/R A)))
(define (box/r x) (lambda () x))
;; we can see that (Boxof/R T1) is a subtype of (Boxof/R T2)
;; if T1 is a subtype of T2 (this is not surprising, since
;; these boxes are similar to any other container, like lists):
(: foo1 : Integer > (Boxof/R Integer))
(define (foo1 b) (box/r b))
(: bar1 : (Boxof/R Number) > Number)
(define (bar1 b) (b))
(test (bar1 (foo1 123)) => 123)
;; a type for a "writeonly" box
(definetype (Boxof/W A) = (A > Void))
;; Boxof/W constructor
(: box/w : (All (A) A > (Boxof/W A)))
(define (box/w x) (lambda (new) (set! x new)))
;; in contrast to the above, (Boxof/W T1) is a subtype of
;; (Boxof/W T2) if T2 is a subtype of T1, *not* the other way
;; (and note how this is related to A being on the *left* side
;; of the arrow in the `Boxof/W' type):
(: foo2 : Number > (Boxof/W Number))
(define (foo2 b) (box/w b))
(: bar2 : (Boxof/W Integer) Integer > Void)
(define (bar2 b new) (b new))
(test (bar2 (foo2 123) 456))
;; combining the above two into a type for a "read/write" box
(definetype (Boxof/RW A) = (A > A))
;; Boxof/RW constructor
(: box/rw : (All (A) A > (Boxof/RW A)))
(define (box/rw x) (lambda (new) (let ([old x]) (set! x new) old)))
;; this combines the above two: `A' appears on both sides of the
;; arrow, so (Boxof/RW T1) is a subtype of (Boxof/RW T2) if T1
;; is a subtype of T2 (because there's an A on the right) *and*
;; if T2 is a subtype of T1 (because there's another A on the
;; left)  and that can happen only when T1 and T2 are the same
;; type. So this is a type error:
;; (: foo3 : Integer > (Boxof/RW Integer))
;; (define (foo3 b) (box/rw b))
;; (: bar3 : (Boxof/RW Number) Number > Number)
;; (define (bar3 b new) (b new))
;; (test (bar3 (foo3 123) 456) => 123)
;; ** Expected (Number > Number), but got (Integer > Integer)
;; And this a type error too:
;; (: foo3 : Number > (Boxof/RW Number))
;; (define (foo3 b) (box/rw b))
;; (: bar3 : (Boxof/RW Integer) Integer > Integer)
;; (define (bar3 b new) (b new))
;; (test (bar3 (foo3 123) 456) => 123)
;; ** Expected (Integer > Integer), but got (Number > Number)
;; The two types must be the same for this to work:
(: foo3 : Integer > (Boxof/RW Integer))
(define (foo3 b) (box/rw b))
(: bar3 : (Boxof/RW Integer) Integer > Integer)
(define (bar3 b new) (b new))
(test (bar3 (foo3 123) 456) => 123)
Implementing a Circular Environment
We now use this to implement rec
in the following way:

Change environments so that instead of values they hold boxes of values:
(Boxof VAL)
instead ofVAL
, and wheneverlookup
is used, the resulting boxed value is unboxed, 
In the
WRec
case, create the new environment with some temporary binding for the identifier — any value will do since it should not be used (when named expressions are alwaysfun
expressions), 
Evaluate the expression in the new environment,

Change the binding of the identifier (the box) to the result of this evaluation.
The resulting definition is:
;; 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)
(let ([newcell (box (NumV 42))])
(let ([newenv (Extend id newcell restenv)])
(let ([value (eval expr newenv)])
(setbox! newcell value)
newenv))))
Racket has another let
relative for such cases of multiplenested
let
s — let*
. This form is a derived form — it is defined as a
shorthand for using nested let
s. The above is therefore exactly the
same as this code:
;; 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)
(let* ([newcell (box (NumV 42))]
[newenv (Extend id newcell restenv)]
[value (eval expr newenv)])
(setbox! newcell value)
newenv))
This let*
form can be read almost as a C/Javaish kind of code:
new_cell = new NumV(42);
new_env = Extend(id, new_cell, rest_env);
value = eval(expr, new_env);
*new_cell = value;
return new_env;
}
The code can be simpler if we fold the evaluation into the setbox!
(since value
is used just there), and if use lookup
to do the
mutation — since this way there is no need to hold onto the box. This
is a bit more expensive, but since the binding is guaranteed to be the
first one in the environment, the addition is just one quick step. The
only binding that we need is the one for the new environment, which we
can do as an internal definition, leaving us with:
(define (extendrec id expr restenv)
(define newenv (Extend id (box (NumV 42)) restenv))
(setbox! (lookup id newenv) (eval expr newenv))
newenv)
A complete rehacked version of FLANG with a rec
binding follows. We
can’t test rec
easily since we have no conditionals, but you can at
least verify that
is an infinite loop.
flangbox.rkt^{ D }#lang pl
(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 (or 'with 'rec) more)
(match sexpr
[(list 'with (list (symbol: name) named) body)
(With name (parsesexpr named) (parsesexpr body))]
[(list 'rec (list (symbol: name) named) body)
(WRec name (parsesexpr named) (parsesexpr body))]
[(cons x more)
(error 'parsesexpr "bad `~s' syntax in ~s" x 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 ENV
[EmptyEnv]
[Extend Symbol (Boxof VAL) ENV])
(definetype VAL
[NumV Number]
[FunV Symbol FLANG ENV])
(: lookup : Symbol ENV > (Boxof VAL))
;; lookup a symbol in an environment, return its value or throw an
;; error if it isn't bound
(define (lookup name env)
(cases env
[(EmptyEnv) (error 'lookup "no binding for ~s" name)]
[(Extend id boxedval restenv)
(if (eq? id name) boxedval (lookup name restenv))]))
(: 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)
(define newenv (Extend id (box (NumV 42)) restenv))
(setbox! (lookup id newenv) (eval expr newenv))
newenv)
(: 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 (box (eval namedexpr env)) env))]
[(WRec boundid namedexpr boundbody)
(eval boundbody
(extendrec boundid namedexpr env))]
[(Id name) (unbox (lookup name env))]
[(Fun boundid boundbody)
(FunV boundid boundbody env)]
[(Call funexpr argexpr)
(define fval (eval funexpr env))
(cases fval
[(FunV boundid boundbody fenv)
(eval boundbody
(Extend boundid (box (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)