Oxcart

Try it online @ catseye.tc | Wiki entry @ esolangs.org | See also: Carriage ∘ Equipage ∘ Wagon

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While desigining Wagon the topic of continuations briefly came up. I didn't, at the time, think that thinking in terms of continuations would make designing Wagon any easier. But I did remark that a continuation-passing concatenative language sounded like an interesting thing in its own right.

After Wagon was finished, I began thinking about how one would make continuation-passing concatenative language, but I immediately hit a wall: how do you compose two functions written in continuation-passing style?

So I sat down and worked it out. Maybe you can do it, I thought, if you adopt a non-standard formulation of function composition.

If conventional function composition is defined as

(f ∘ g)(x) = g(f(x))

then, rather arbitrarily picking the symbol ⊛ to denote it, composition of CPS functions can be defined as

(f ⊛ g)(x, κ) = f(x, λs. g(s, κ))

or alternately,

(f ⊛ g) = λ(x, κ). f(x, λs. g(s, κ))

The question that remains is whether this is a workable substitute for conventional function composition in a concatenative language.

This question has two parts: whether it's algebraically valid, and whether it's useful for writing programs with.

Algebraic properties of ⊛

The first part. Functions form a monoid under composition; there is an identity element (the identity function):

e(x) = x

and this is an identity because

(e ∘ f)(x) = f(e(x)) = f(x)
(f ∘ e)(x) = e(f(x)) = f(x)

and composition is associative:

((f ∘ g) ∘ h) = (f ∘ (g ∘ h))

because

((f ∘ g) ∘ h) = (f ∘ (g ∘ h))
(g(f(x)) ∘ h) = (f ∘ (h(g(x)))
(h(g(f(x))) = (h(g(f(x))))

Can we devise an identity CPS function? I think it might be:

ι(x, κ) = κ(x)

and this is an identity because

(ι ⊛ f)(x, κ) = ι(x, λs. f(s, κ)) = (λs. f(s, κ))(x) = f(x, κ)
(f ⊛ ι)(x, κ) = f(x, λs. ι(s, κ)) = f(x, λs. κ(s))) = f(x, κ)

And is ⊕ associative? Well, let's try expanding it:

((f ⊛ g) ⊛ h)

replace (f ⊛ g) with λ(x, κ). f(x, λs. g(s, κ)):

(λ(x, κ). f(x, λs. g(s, κ)) ⊛ h)

replace (N ⊛ h) with λ(x, j). N(x, λt. h(t, j))
where N = (λ(x, κ). f(x, λs. g(s, κ)))
to get

λ(x, j). (λ(x, κ). f(x, λs. g(s, κ)))(x, λt. h(t, j))

Now reduce (λ(x, κ). f(x, λs. g(s, κ)))(x, λt. h(t, j))
by replacing in the lambda body
x with x and
κ with λt. h(t, j)
to get
f(x, λs. g(s, λt. h(t, j)))

and the whole thing reads

λ(x, j). f(x, λs. g(s, λt. h(t, j)))

which looks reasonable.

Versus:

(f ⊛ (g ⊛ h))

replace (g ⊛ h) with λ(x, κ). g(x, λs. h(s, κ)):

(f ⊛ λ(x, κ). g(x, λs. h(s, κ)))

replace (f ⊛ N) with λ(x, j). f(x, λt. N(t, j))
where N = (λ(x, κ). g(x, λs. h(s, κ)))
to get

λ(x, j). f(x, λt. (λ(x, κ). g(x, λs. h(s, κ)))(t, j))

Now reduce (λ(x, κ). g(x, λs. h(s, κ)))(t, j)
by replacing in the lambda body
x with t and
κ with j
to get
g(t, λs. h(s, j))

and the whole thing reads

λ(x, j). f(x, λt. g(t, λs. h(s, j)))

Yes! It looks like it is!

A concatenative language with ⊛

Now the second part. This requires us to actually try to define some kind of concatenative language around this formulation of composition, and see what kind of programs we can write in it.

Like Carriage and Equipage and Wagon, this will be a "purely concatenative language": the entire program is a single string of sequentially concatenated symbols, and each symbol represents a function, and the functions are sequentially composed in the same manner the symbols are concatenated. More to the point, you don't get to name anything or to nest anything inside anything else.

Unlike Wagon we won't be concerned with expressing control outside of the program state. Indeed, first-class continuations are a way to reify control as data, so we'll happily make them part of the data store.

I'm sure we could get away with having a single stack for the store, like most concatenative languages, but to make things easier (maybe) let's deviate slightly. A store, in Oxcart, is a tape of stacks. That is, it's an unbounded array of stacks, plus an index into that array. The index is initially 0 but can be changed; the stack that it points to is referred to as "the current stack", and most operations operate on the current stack.

Each stack is strictly FIFO and initially empty, and each stack cell can hold either an int or a continuation. Ints are generally assumed to be 64 bits in this day and age, but it pays to be cautious.

Basic operations

-> Tests for functionality "Evaluate Oxcart Program"

-> Functionality "Evaluate Oxcart Program" is implemented by
-> shell command "bin/oxcart %(test-body-file)"

The instruction 0 pushes a zero onto the current stack.

| 0
= > 0:[0]

Whitespace is a no-op.

|       
=

These demonstrate how Oxcart stores are represented on output by the reference implementation: the current stack is indicated by >, followed by its index, then :, then its contents, top-to-bottom. But only stacks that are non-empty are output.

The instruction ^ (resp. v) pops a value from the current stack, increments (resp. decrements) it, and pushes the result back onto the current stack.

| 0^^^0vv
= > 0:[-2,3]

The instruction : pops a value from the current stack and pushes two copies of the value back on the stack.

| 0^^^^^^^^:^
= > 0:[9,8]

The instruction $ pops a value from the current stack and discards it.

| 0^^^^^$
=

The instruction \\ pops the top two values, swaps them, and pushes them back on the stack.

| 0^^^^^^^^0^\0^^
= > 0:[2,8,1]

Navigating the stacks

The instruction < (resp >) moves one space left (resp. right) on the tape, changing which stack is the current stack.

| 0^^^^<0^^^^^^^^<0^^^^^^^^^^>
=  -2:[10]
= >-1:[8]
=   0:[4]

The instruction ( (resp )) pops a value off the current stack, moves one space left (resp. right) on the tape, and pushes the value onto the new current stack.

| 0^^^^<0^^^^^^^^(0^^^^^^^^^^)
=  -2:[8]
= >-1:[10]
=   0:[4]

The instruction ' (apostrophe) pops a first value off the stack, then a second value. It then sets the tape position to the first value, and pushes the second value back on the (probably newly current) stack.

| <0^^^0^^^^^0^'
=  -1:[3]
= > 1:[5]

You can of course push a dummy value, then discard it after moving it, if all you want to do is change to a different stack.

| <<<<<<00'$ 0^
= > 0:[1]

The instruction Y pops a first value off the stack, then a second value off the stack.

If the first value is non-zero, nothing else in particular happens and evaluation continues as usual.

| 0^^0^0^Y0^^^
= > 0:[3,2]

But if the first value is zero, the second value is added to the tape position (negative values go left, positive values go right).

| 0^^0^0Y0^^^
=   0:[2]
= > 1:[3]

| 0^^0v0Y0^^^
= >-1:[3]
=   0:[2]

Operations involving continuations

The instruction S pushes the current continuation onto the stack. Note that continuations don't have a defined representation other than #k.

| S
= > 0:[#k]

The instruction % pops a first value off the stack, then a second value off the stack.

If the first value is zero, nothing happens and evaluation continues as usual.

| S0%
=

But if the first value is non-zero, it replaces the current continuation with the second value, and continues with that continuation.

| 0^^^0S0^%
= > 0:[3]

In the preceding example, when % is evaluated, the 1 pushed by the 0^ just before the %, and the continuation pushed by S, are popped off the stack (leaving 0 and 3 on the stack.) The 1 is judged to be non-zero, so the continuation pushed by S is continued. That continuation represents the remainder of the program that consists of 0^%. So a 1 is pushed onto the stack and % is evaluated again. But this time % gets a 1 and a 0, which is not a continuation, so things continue as usual. The result is only the initial 3 on the stack.

Infinite loop

So we want to write an infinite loop. In high-level terms, we need to save the current continuation in a value k. (Note that when we continue k, we'll end up back at this point.) Then we want to continue k. (Note that, since we end up back at that point noted previously, we never get to this point.)

We can write this in Oxcart as:

S:0^%

(We don't write this as a Falderal test, because we want all our tests to terminate. But it is provided as a discrete program in the eg/ directory, if you want to run it.)

Controlled loop

So we want to write a loop that terminates. Say we want to generate the numbers from 10 down to 0. In high-level terms, we set a value n to 10, and save the current continuation as k. Then we make a copy of n and decrement it to obtain n'. Then we make a copy of n' and test if it's zero. If it is, we're done. If not, we continue k.

We can write this in Oxcart as:

• move left
• push 10 on stack
• move right
• push current continuation on stack
• duplicate
• move left
• duplicate
• decrement
• duplicate
• transfer right
• continue conditionally

Or, as an actual Oxcart program:

| <0^^^^^^^^^^>S:<:v:)%
=  -1:[0,1,2,3,4,5,6,7,8,9,10]
= > 0:[#k]

While loop?

So, while we've demonstrated it's possible to write a controlled loop, it is in fact a "repeat" (or "do") type loop, where the loop body is always executed at least once. What about a "while" type loop, where the loop body might not be executed at all, if the loop condition isn't true when the loop starts?

You may have noticed that the "current continuation" is a very palpable concept in Oxcart; using the infinite loop program to illustrate, it is almost as if concatenating the program symbols results in a program structured like this:

S→:→0→^→%→■

where each → is a continuation, and ■ is HALT, and execution happens by executing one instruction, then just following the attached arrow to get to the next instruction to execute. An instruction like S has the effect of pushing the arrow (and, virtually, everything that follows it) onto the stack, and an instruction like % also has an arrow attached to it, but that arrow is ignored; an arrow popped off the stack is followed instead.

But one implication of this is that an Oxcart program can't access any continuation it hasn't already "seen", i.e. any continuations that it might encounter down the line, in the future. In more pedestrian terms, you can't denote a forward jump.

And that means we can't write a "while" loop in the usual manner.

But perhaps we can write one in a slightly unconventional manner.

The idea is this: the body of the loop is executed at least once, but it is executed in a context where it has no effect on anything we care about.

This might not work, but let's try to work it out.

So we want to write a "while" loop. Say we have an n on the stack, and we want to loop n times, and n might be zero.

In high-level terms, we first move to a "junk stack" and place a "junk n" on it.

Then, we save the current continuation as k.

We test if n is zero. If it is, we switch to a junk stack.

Then, assuming we're on the real stack, we make a copy of n and decrement it to obtain n'. Then we make a copy of n' and test if it's zero. If it is, we're done. If not, we continue k.

But, assuming we're on the junk stack, the above becomes: we make a copy of junk n and decrement it to obtain junk n'. Then we make a copy of junk n' and test if it's zero. If it is, we're done. If not, we continue k.

This suggests our initial junk n should be 1.

The problem is that we want to switch back from the junk stack to the real stack if previously we were on the junk stack. (This is what preciptated adding the ' instruction to the language.)

Can we can write this in Oxcart?

• transfer left (to move n to the data stack, -1)
• move left (to junk stack, -2)
• push 1 on stack
• reset to the main stack
• push current continuation on stack
• duplicate
• move left
• duplicate
• pop and if value is zero move one stack to the left
• duplicate
• decrement
• duplicate
• transfer value to the main stack (this is the test value)
• continue conditionally
• pop and discard the original pushed continuation

Is this it? With n=5:

| 0^^^^^
| (<0^00'$S:<:0v\Y:v:0'%$
=  -2:[1]
=  -1:[0,1,2,3,4,5]

And with n=0:

| 0
| (<0^00'$S:<:0v\Y:v:0'%$
=  -2:[0,1]
=  -1:[0]

Hooray! I think we just built a while loop. One might need one junk stack per while loop, but one would only have a finite and fixed number of while loops in any given program anyway.

I have not shown that it is possible to nest while loops. Offhand, it seems plausible. It may be slightly complicated, in that the top-level while loop must junk its first iteration only once, but the inner while loop needs to junk its first iteration many times. So there might need to be some code that resets the inner loop's junk stack to a safely junkable state. But it definitely feels more like it's doable, than like it's insurmountable.

Minimality of Oxcart

Oxcart is not a minimal language. It defines operations that are redundant with other operations.

One could define a "Core Oxcart" that omits the following operations:

<>\\

Because < and > can be thought of as just shorthands for 0v0^Y and 0^0^Y.

And \\ can be implemented using <, (, ), and >, as follows.

| 0^0^^
= > 0:[2,1]

| 0^0^^)<(>>(<)
= > 0:[1,2]

Or in fact you can build a "rotate" of arbitrary finite depth with those operations.

It's possible Core Oxcart could omit other operations, too, if they turn out to be not critical for showing that it is Turing-complete. In particular, the ' operation is very powerful, rather repulsively so; it's the only operation that lets you address tape cells in an absolute fashion. You might be able to use it where you would otherwise use (). It would probably be nicer to replace it with something that also works relatively, like <, (, ), >, and Y do.

But the goal of Oxcart was not to make a "nice esolang", and in fact the whole "tape of stacks" thing was entirely a secondary design choice; the main goal was to show that a contiuation-passing concatenative language was viable, and I think it achieved that goal. Making a CPS concatenative language which is also a "nice esolang" can be saved for future work.

Bumpy trails!
Chris Pressey
London, UK
October 28th, 2019