Lanthorn

Version 1.0 | Entry @ catseye.tc
| See also: Lariat

• Iphigeneia

When I first came across a explanation of how letrec works, it was
in terms of updating references: each of the names is bound to a cell,
and when the thing that name refers to is eventually defined, that cell
is updated with that thing.

My reaction to this was ugh. I mean, sure, it works, but in the
context of functional programming, such an imperative description is
really unsatisfying.

So, I present here a tiny, eager, purely functional language, christened
Lanthorn, whose sole purpose is to host a demonstration of how letrec
can be written as syntactic sugar over let in a purely functional way.

The transformation is unobtrusive in that it doesn't make any changes in
the body of the letrec. The resulting code is not, however, intended
to be efficient.

Since the language is simple enough and conventional enough that you
can probably guess what the programs mean, let's leave the description
of the language until Appendix A, and go straight into
describing the transformation.

Desugaring

-> Tests for functionality "Desugar Lanthorn Program"

Basically, what we want to do, is take this...

letrec
    odd  = fun(x) -> if eq(x, 0) then false else even(sub(x, 1))
    even = fun(x) -> if eq(x, 0) then true else odd(sub(x, 1))
in
    even(6)

...and turn it into this.

let
    odd0  = fun(x, odd1, even1) ->
                 let
                     odd = fun(x) -> odd1(x, odd1, even1)
                     even = fun(x) -> even1(x, odd1, even1)
                 in
                     if eq(x, 0) then false else even(sub(x, 1)))
    even0 = fun(x, odd1, even1) ->
                 let
                     odd = fun(x) -> odd1(x, odd1, even1)
                     even = fun(x) -> even1(x, odd1, even1)
                 in
                     if eq(x, 0) then true else odd(sub(x, 1)))
    odd   = fun(x) -> odd0(x, odd0, even0)
    even  = fun(x) -> even0(x, odd0, even0)
in
    even(6)

Our evaluator implements this transformation in the
Language.Lanthorn.LetRec module.
Here is what it produces. Note there is a bit of name-mangling
added, compared to the hand-written expansion above.

letrec
    odd  = fun(x) -> if eq(x, 0) then false else even(sub(x, 1))
    even = fun(x) -> if eq(x, 0) then true else odd(sub(x, 1))
in
    even(6)
=> let
=>   odd$0 = fun(x, odd$1, even$1) -> let
=>       odd = fun(x$1) -> odd$1(x$1, odd$1, even$1)
=>       even = fun(x$1) -> even$1(x$1, odd$1, even$1)
=>     in
=>       if eq(x, 0) then false else even(sub(x, 1))
=>   even$0 = fun(x, odd$1, even$1) -> let
=>       odd = fun(x$1) -> odd$1(x$1, odd$1, even$1)
=>       even = fun(x$1) -> even$1(x$1, odd$1, even$1)
=>     in
=>       if eq(x, 0) then true else odd(sub(x, 1))
=>   odd = fun(x) -> odd$0(x, odd$0, even$0)
=>   even = fun(x) -> even$0(x, odd$0, even$0)
=> in
=>   even(6)

In English, it adds a number of extra parameters to each function in
the set of bindings. Specifically, it adds one parameter for each
of the bindings. It then sets up some bindings inside each function
so that the function uses these parameters for the recursive calls
it makes. It also sets up some bindings outside of these functions
to that the body of the letrec sees functions with the original
parameters they had, hiding all these extra parameters.

Related Work

Xavier Pinho has written up an alternative way of transforming letrec
into let, using surjective pairing and the Y combinator, in
an issue on the Lanthorn project on GitHub.

Appendix A

Basic Syntax of Lanthorn

-> Tests for functionality "Pretty-print Lanthorn Program"

Function application, numeric literals, string literals.

add(1, 2)
=> add(1, 2)

Name binding (let) and name reference.

let a = 1
    b = 1
    in zed(a, b)
=> let
=>   a = 1
=>   b = 1
=> in
=>   zed(a, b)

The character $ may not appear in user-supplied names.

let
  a$b = 1
in
  zed(a$b)
?> unexpected "$"

Conditional by boolean (if).

if gt(a, b) then a else b
=> if gt(a, b) then a else b

Function values.

let up = fun(x) -> add(x, 1) in up(5)
=> let
=>   up = fun(x) -> add(x, 1)
=> in
=>   up(5)

Basic Semantics of Lanthorn

-> Tests for functionality "Evaluate Lanthorn Program"

1
===> 1

if true then 5 else 6
===> 5

let a = 2 in a
===> 2

Basic functions.

let r = fun(x) -> 77 in r(1)
===> 77

let r = fun(x) -> x in r(66)
===> 66

let is like Scheme's let* or Standard ML's let:
later bindings can see earlier bindings.

let
  p = 99
  r = fun(x) -> p
in
  r(66)
===> 99

Note that depicting a function is implementation-dependent.

fun(x) -> x
===> <<function>>

Can't shadow a binding in let.

let a = 1 in let a = 2 in a
???> Already defined: a

Can't shadow a binding in the formals of a fun.

let r = fun(x, x) -> x in r(10, 10)
???> Already defined: x

let r = fun(x) -> let x = 3 in x in r(10)
???> Already defined: x

letrec

Basic usage of letrec.

letrec
    oddp  = fun(x) -> if eq(x, 0) then false else evenp(sub(x, 1))
    evenp = fun(x) -> if eq(x, 0) then true else oddp(sub(x, 1))
in
    evenp(6)
===> true

letrec
    oddp  = fun(x) -> if eq(x, 0) then false else evenp(sub(x, 1))
    evenp = fun(x) -> if eq(x, 0) then true else oddp(sub(x, 1))
in
    evenp(5)
===> false

letrec nested inside an if inside a function definition in an arm of
another letrec.

letrec
    facto = fun(n) -> if eq(n, 1) then 1 else
        letrec
            oddp  = fun(x) -> if eq(x, 0) then false else evenp(sub(x, 1))
            evenp = fun(x) -> if eq(x, 0) then true else oddp(sub(x, 1))
        in
            if oddp(n) then
                mul(n, facto(sub(n, 1)))
            else
                facto(sub(n, 1))
in
    facto(8)
===> 105

letrec nested in the body of another letrec.

letrec
    oddp  = fun(x) -> if eq(x, 0) then false else evenp(sub(x, 1))
    evenp = fun(x) -> if eq(x, 0) then true else oddp(sub(x, 1))
in
    letrec facto = fun(n) ->
        if eq(n, 1) then
            1
        else if oddp(n) then
            mul(n, facto(sub(n, 1)))
        else
            facto(sub(n, 1))
in
    facto(8)
===> 105

Nested letrec, nested right in the arm of another letrec. Currently,
this is an error, because the inner scope cannot "see" the outer letrec.
Though I'm not yet convinced of what the most reasonable behaviour is here.

letrec
    facto =
        letrec
            oddp  = fun(x) -> if eq(x, 0) then false else evenp(sub(x, 1))
            evenp = fun(x) -> if eq(x, 0) then true else oddp(sub(x, 1))
        in
            fun(n) -> if eq(n, 1) then 1 else
                if oddp(n) then
                    mul(n, facto(sub(n, 1)))
                else
                    facto(sub(n, 1))
in
    facto(8)
???> Not in scope: facto

letrec nested inside a function definition inside an arm of a plain let.

let
    factoo = fun(f, n) ->
        letrec
            oddp  = fun(x) -> if eq(x, 0) then false else evenp(sub(x, 1))
            evenp = fun(x) -> if eq(x, 0) then true else oddp(sub(x, 1))
        in
            if eq(n, 1) then 1 else
                if oddp(n) then
                    mul(n, f(f, sub(n, 1)))
                else
                    f(f, sub(n, 1))
in
    factoo(factoo, 7)
===> 105

letrec nested inside body of a plain let.

let
    factopen = fun(f, n) -> if eq(n, 1) then 1 else mul(n, f(f, sub(n, 1)))
    target = 7
in
    letrec
        oddp  = fun(x) -> if eq(x, 0) then false else evenp(sub(x, 1))
        evenp = fun(x) -> if eq(x, 0) then true else oddp(sub(x, 1))
    in
        if oddp(target) then factopen(factopen, target) else 0
===> 5040

letrec works on functions that have more than one argument.

letrec
    oddsump  = fun(x,y,z) -> if eq(add(x, add(y, z)), add(y, z)) then false else evensump(sub(x, 1), y, z)
    evensump = fun(x,y,z) -> if eq(add(x, add(y, z)), add(y, z)) then true else oddsump(sub(x, 1), y, z)
in
    evensump(5,3,1)
===> false

letrec
    oddsump  = fun(x,y,z) -> if eq(add(x, add(y, z)), add(y, z)) then false else evensump(sub(x, 1), y, z)
    evensump = fun(x,y,z) -> if eq(add(x, add(y, z)), add(y, z)) then true else oddsump(sub(x, 1), y, z)
in
    evensump(6,3,1)
===> true

letrec works on functions which use different argument names.

letrec
    oddsump  = fun(x,y,z) -> if eq(add(x, add(y, z)), add(y, z)) then false else evensump(sub(x, 1), y, z)
    evensump = fun(p,q,r) -> if eq(add(p, add(q, r)), add(q, r)) then true else oddsump(sub(p, 1), q, r)
in
    evensump(5,3,1)
===> false

letrec
    oddsump  = fun(x,y,z) -> if eq(add(x, add(y, z)), add(y, z)) then false else evensump(sub(x, 1), y, z)
    evensump = fun(p,q,r) -> if eq(add(p, add(q, r)), add(q, r)) then true else oddsump(sub(p, 1), q, r)
in
    evensump(6,3,1)
===> true

letrec works on functions that have differing numbers of arguments.

letrec
    oddsump  = fun(x,y,z) -> if eq(add(x, add(y, z)), add(y, z)) then false else evensump(sub(x, 1), add(y, z))
    evensump = fun(p,q)   -> if eq(add(p, q), q) then true else oddsump(sub(p, 1), 1, sub(q, 1))
in
    oddsump(5,3,1)
===> true

Properties of the letrec transformation

-> Tests for functionality "Desugar Lanthorn Program"

When a letrec is desugared, the generated functions have argument
names that are based on the original argument names. Also, the
innermost lets bind the plain names to functions with the same arity
as the original functions.

letrec
    oddsump  = fun(x,y,z) -> if eq(add(x, add(y, z)), add(y, z)) then false else evensump(sub(x, 1), y, z)
    evensump = fun(x,y,z) -> if eq(add(x, add(y, z)), add(y, z)) then true else oddsump(sub(x, 1), y, z)
in
    evensump(5,3,1)
=> let
=>   oddsump$0 = fun(x, y, z, oddsump$1, evensump$1) -> let
=>       oddsump = fun(x$1, y$1, z$1) -> oddsump$1(x$1, y$1, z$1, oddsump$1, evensump$1)
=>       evensump = fun(x$1, y$1, z$1) -> evensump$1(x$1, y$1, z$1, oddsump$1, evensump$1)
=>     in
=>       if eq(add(x, add(y, z)), add(y, z)) then false else evensump(sub(x, 1), y, z)
=>   evensump$0 = fun(x, y, z, oddsump$1, evensump$1) -> let
=>       oddsump = fun(x$1, y$1, z$1) -> oddsump$1(x$1, y$1, z$1, oddsump$1, evensump$1)
=>       evensump = fun(x$1, y$1, z$1) -> evensump$1(x$1, y$1, z$1, oddsump$1, evensump$1)
=>     in
=>       if eq(add(x, add(y, z)), add(y, z)) then true else oddsump(sub(x, 1), y, z)
=>   oddsump = fun(x, y, z) -> oddsump$0(x, y, z, oddsump$0, evensump$0)
=>   evensump = fun(x, y, z) -> evensump$0(x, y, z, oddsump$0, evensump$0)
=> in
=>   evensump(5, 3, 1)

The transformation mangles names that it generates so that they never
shadow names that appear in the user's program. (Names containing $ may
not appear in a user-supplied program.)

let
    odd0 = fun(a, b, c) -> a
in
    letrec
        odd  = fun(x) -> if eq(x, 0) then false else even(sub(x, 1))
        even = fun(x) -> if eq(x, 0) then true else odd(sub(x, 1))
    in
        even(6)
=> let
=>   odd0 = fun(a, b, c) -> a
=> in
=>   let
=>     odd$0 = fun(x, odd$1, even$1) -> let
=>         odd = fun(x$1) -> odd$1(x$1, odd$1, even$1)
=>         even = fun(x$1) -> even$1(x$1, odd$1, even$1)
=>       in
=>         if eq(x, 0) then false else even(sub(x, 1))
=>     even$0 = fun(x, odd$1, even$1) -> let
=>         odd = fun(x$1) -> odd$1(x$1, odd$1, even$1)
=>         even = fun(x$1) -> even$1(x$1, odd$1, even$1)
=>       in
=>         if eq(x, 0) then true else odd(sub(x, 1))
=>     odd = fun(x) -> odd$0(x, odd$0, even$0)
=>     even = fun(x) -> even$0(x, odd$0, even$0)
=>   in
=>     even(6)

-> Tests for functionality "Evaluate Lanthorn Program"

letrec
    odd  = fun(x) -> if eq(x, 0) then false else even(sub(x, 1))
    odd0 = fun(a, b, c) -> a
    even = fun(x) -> if eq(x, 0) then true else odd(sub(x, 1))
in
    even(6)
===> true

You might think that instead of mangling names, we could just allow shadowing
in the language. But that by itself doesn't solve our problem, since you
could still say something like the following. The letrec desugaring would
have to be more aware of how it constructs names, at any rate, in order to
avoid the conflict here. And mangling is the simplest way to do that.

-> Tests for functionality "Desugar Lanthorn Program"

letrec
    odd  = fun(x) -> if eq(x, 0) then false else even(sub(x, 1))
    odd0 = fun(a, b, c) -> a
    even = fun(x) -> if eq(x, 0) then true else odd(sub(x, 1))
in
    even(6)
=> let
=>   odd$0 = fun(x, odd$1, odd0$1, even$1) -> let
=>       odd = fun(x$1) -> odd$1(x$1, odd$1, odd0$1, even$1)
=>       odd0 = fun(a$1, b$1, c$1) -> odd0$1(a$1, b$1, c$1, odd$1, odd0$1, even$1)
=>       even = fun(x$1) -> even$1(x$1, odd$1, odd0$1, even$1)
=>     in
=>       if eq(x, 0) then false else even(sub(x, 1))
=>   odd0$0 = fun(a, b, c, odd$1, odd0$1, even$1) -> let
=>       odd = fun(x$1) -> odd$1(x$1, odd$1, odd0$1, even$1)
=>       odd0 = fun(a$1, b$1, c$1) -> odd0$1(a$1, b$1, c$1, odd$1, odd0$1, even$1)
=>       even = fun(x$1) -> even$1(x$1, odd$1, odd0$1, even$1)
=>     in
=>       a
=>   even$0 = fun(x, odd$1, odd0$1, even$1) -> let
=>       odd = fun(x$1) -> odd$1(x$1, odd$1, odd0$1, even$1)
=>       odd0 = fun(a$1, b$1, c$1) -> odd0$1(a$1, b$1, c$1, odd$1, odd0$1, even$1)
=>       even = fun(x$1) -> even$1(x$1, odd$1, odd0$1, even$1)
=>     in
=>       if eq(x, 0) then true else odd(sub(x, 1))
=>   odd = fun(x) -> odd$0(x, odd$0, odd0$0, even$0)
=>   odd0 = fun(a, b, c) -> odd0$0(a, b, c, odd$0, odd0$0, even$0)
=>   even = fun(x) -> even$0(x, odd$0, odd0$0, even$0)
=> in
=>   even(6)

-> Tests for functionality "Evaluate Lanthorn Program"

let
    odd0 = fun(a, b, c) -> a
in
    letrec
        odd  = fun(x) -> if eq(x, 0) then false else even(sub(x, 1))
        even = fun(x) -> if eq(x, 0) then true else odd(sub(x, 1))
    in
        even(6)
===> true

Note that there is probably a case where a letrec nested another letrec, and
which shadows variables of the enclosing letrec, produces a less readable
error message about shadowing, because it mentions the mangled names; but
I can live with that for now.