Lecture #13, Tuesday, February 18th
===================================
 Alternative Church Encoding
 Recursive Environments
 Recursion: Racket's `letrec`
 Implementing Recursion using `letrec`

# Alternative Church Encoding
Finally, note that this is just one way to encode things  other
encodings are possible. One alternative encoding is in the following
code  it uses a list of `N` falses as the encoding for `N`. This
encoding makes it easier to `add1` (just `cons` another `#f`), and to
`sub1` (simply `cdr`). The tradeoff is that some arithmetics operations
becomes more complicated, for example, the definition of `+` requires
the fixpoint combinator. (As expected, some people want to see what can
we do with a language without recursion, so they don't like jumping to Y
too fast.)
;;; <<>>
;; An alternative "Church" encoding: use lists to encode numbers
#lang pl schlac
(define identity (lambda (x) x))
;; Booleans (same as before)
(define #t (lambda (x y) x))
(define #f (lambda (x y) y))
(define if (lambda (c t e) (c t e))) ; not really needed
(test (>bool (if #t #f #t))
=> '#f)
(test (>bool (if #f ((lambda (x) (x x)) (lambda (x) (x x))) #t))
=> '#t)
(define and (lambda (a b) (a b a)))
(define or (lambda (a b) (a a b)))
(define not (lambda (a x y) (a y x)))
(test (>bool (and #f #f)) => '#f)
(test (>bool (and #t #f)) => '#f)
(test (>bool (and #f #t)) => '#f)
(test (>bool (and #t #t)) => '#t)
(test (>bool (or #f #f)) => '#f)
(test (>bool (or #t #f)) => '#t)
(test (>bool (or #f #t)) => '#t)
(test (>bool (or #t #t)) => '#t)
(test (>bool (not #f)) => '#t)
(test (>bool (not #t)) => '#f)
;; Lists (same as before)
(define cons (lambda (x y s) (s x y)))
(define car (lambda (x) (x #t)))
(define cdr (lambda (x) (x #f)))
(define 1st car)
(define 2nd (lambda (l) (car (cdr l))))
(define 3rd (lambda (l) (car (cdr (cdr l)))))
(define 4th (lambda (l) (car (cdr (cdr (cdr l))))))
(define 5th (lambda (l) (car (cdr (cdr (cdr (cdr l)))))))
(define null (lambda (s) #t))
(define null? (lambda (x) (x (lambda (x y) #f))))
;; Natural numbers (alternate encoding)
(define 0 identity)
(define add1 (lambda (n) (cons #f n)))
(define zero? car) ; tricky
(define sub1 cdr) ; this becomes very simple
;; Note that we could have used something more straightforward:
;; (define 0 null)
;; (define add1 (lambda (n) (cons #t n))) ; cons anything
;; (define zero? null?)
;; (define sub1 (lambda (l) (if (zero? l) l (cdr l))))
(define 1 (add1 0))
(define 2 (add1 1))
(define 3 (add1 2))
(define 4 (add1 3))
(define 5 (add1 4))
(test (>nat* (add1 (add1 5))) => '7)
(test (>nat* (sub1 (sub1 (add1 (add1 5))))) => '5)
(test (>bool (and (zero? 0) (not (zero? 3)))) => '#t)
(test (>bool (zero? (sub1 (sub1 (sub1 3))))) => '#t)
;; listofnumbers tests
(define l123 (cons 1 (cons 2 (cons 3 null))))
(test (>listof >nat* l123) => '(1 2 3))
(test (>listof (>listof >nat*) (cons l123 (cons l123 null)))
=> '((1 2 3) (1 2 3)))
;; Recursive functions
(define Y
(lambda (f)
((lambda (x) (x x)) (lambda (x) (f (x x))))))
(rewrite (define/rec f E) => (define f (Y (lambda (f) E))))
;; note that this example is doing something silly now
(define/rec length
(lambda (l)
(if (null? l)
0
(add1 (length (cdr l))))))
(test (>nat* (length l123)) => '3)
;; addition becomes hard since it requires a recursive definition
;; (define/rec +
;; (lambda (m n) (if (zero? n) m (+ (add1 m) (sub1 n)))))
;; (test (>nat* (+ 4 5)) => '9)
;; faster alternative:
(define/rec +
(lambda (m n)
(if (zero? m) n
(if (zero? n) m
(add1 (add1 (+ (sub1 m) (sub1 n))))))))
(test (>nat* (+ 4 5)) => '9)
;; subtraction is similar to addition
;; (define/rec 
;; (lambda (m n) (if (zero? n) m ( (sub1 m) (sub1 n)))))
;; (test (>nat* ( (+ 4 5) 4)) => '5)
;; but this is not "natural subtraction": doesn't work when n>m,
;; because (sub1 0) does not return 0.
;; a solution is like alternative form of +:
(define/rec 
(lambda (m n)
(if (zero? m) 0
(if (zero? n) m
( (sub1 m) (sub1 n))))))
(test (>nat* ( (+ 4 5) 4)) => '5)
(test (>nat* ( 2 5)) => '0)
;; alternatively, could change sub1 above:
;; (define sub1 (lambda (n) (if (zero? n) n (cdr n))))
;; we can do multiplication in a similar way
(define/rec *
(lambda (m n)
(if (zero? m) 0
(+ n (* (sub1 m) n)))))
(test (>nat* (* 4 5)) => '20)
(test (>nat* (+ 4 (* (+ 2 5) 5))) => '39)
;; and the rest of the examples
(define/rec fact
(lambda (x)
(if (zero? x) 1 (* x (fact (sub1 x))))))
(test (>nat* (fact 5)) => '120)
(define/rec fib
(lambda (x)
(if (or (zero? x) (zero? (sub1 x)))
1
(+ (fib (sub1 x)) (fib (sub1 (sub1 x)))))))
(test (>nat* (fib (* 5 2))) => '89)
#
;; Fullyexpanded Fibonacci (note: much shorter than the previous
;; encoding, but see how Y appears twice  two "((lambda" pairs)
(define fib((lambda(f)((lambda(x)(x x))(lambda(x)(f(x x)))))(lambda(
f)(lambda(x)(((((x(lambda(x y)x))(x(lambda(x y)x)))((x(lambda(x y)y)
)(lambda(x y)x)))(lambda(s)(s(lambda(x y)y)(lambda(x)x))))((((lambda
(f)((lambda(x)(x x))(lambda(x)(f(x x))))) (lambda(f)(lambda(m n)((m(
lambda(x y)x))n (((n(lambda(x y)x)) m)(lambda(s)((s (lambda(x y)y))(
lambda(s)((s (lambda(x y)y))((f(m(lambda(x y)y)))(n(lambda(x y)y))))
))))))))(f(x(lambda(x y)y))))(f((x(lambda(x y)y))(lambda(x y)y))))))
)))
#
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 (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 y) x))))
(define cdr (lambda (l) (l #f (lambda (x y) y))))
;; 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 _ignored_ [Variant arg ...] ...)
=> (define Variant
(lambda (arg ...)
(lambda (Variant ...) (Variant arg ...))))
...)
(rewrite (cases value [(Variant 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 ยง11.5]
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)