How is recursive function implemented in functional languages? - recursion

Consider a following recursive function top-level binding:
let f = (a => f a);
(where arg => expr is a lambda expression)
To define f in the global environment, first we have to evaluate value of the f, namely a => f a. But since f isn't defined yet, expression a => f a would cause an error.
How do functional language interpreters/compilers handle this problem?

Related

ml function of type fn : 'a -> 'b

The function:
fn : 'a -> 'b
Now, are there any functions which can be defined and have this type?
There are two possible implementations for that function signature in Standard ML. One employs exceptions, the other recursion:
val raises : 'a -> 'b =
fn a => raise Fail "some error";
(* Infinite looping; satisfies the type signature, *)
(* but won't ever produce anything. *)
val rec loops : 'a -> 'b =
fn a => loops a;
The first solution may be useful for defining a helper function, say bug, which saves a few key strokes:
fun bug msg = raise Fail ("BUG: " ^ msg);
The other solution may be useful for defining server loops or REPLs.
In the Basis library, OS.Process.exit is such a function that returns an unknown generic type 'a:
- OS.Process.exit;
val it = fn : OS.Process.status -> 'a
A small echo REPL with type val repl = fn : unit -> 'a:
fun repl () =
let
val line = TextIO.inputLine TextIO.stdIn
in
case line of
NONE => OS.Process.exit OS.Process.failure
| SOME ":q\n" => OS.Process.exit OS.Process.success
| SOME line => (TextIO.print line ; repl ())
end
You might also find useful this question about the type signature of Haskell's forever function.
I can think of one example:
fun f a = raise Div;
I can think of several:
One that is recursive,
fun f x = f x
Any function that raises exceptions,
fun f x = raise SomeExn
Any function that is mutually recursive, e.g.,
fun f x = g x
and g x = f x
Any function that uses casting (requires specific compiler support, below is for Moscow ML),
fun f x = Obj.magic x
Breaking the type system like this is probably cheating, but unlike all the other functions with this type, this function actually returns something. (In the simplest case, it's the identity function.)
A function that throws if the Collatz conjecture is false, recurses infinitely if true,
fun f x =
let fun loop (i : IntInf.int) =
if collatz i
then loop (i+1)
else raise Collatz
in loop 1 end
which is really just a combination of the first two.
Any function that performs arbitrary I/O and recurses infinitely, e.g.
fun f x = (print "Woohoo!"; f x)
fun repl x =
let val y = read ()
val z = eval y
val _ = print z
in repl x end
One may argue that exceptions and infinite recursion represent the same theoretical value ⊥ (bottom) meaning "no result", although since you can catch exceptions and not infinitely recursive functions, you may also argue they're different.
If you restrict yourself to pure functions (e.g. no printing or exceptions) and only Standard ML (and not compiler-specific features) and you think of the mutually recursive cases as functionally equivalent in spite of their different recursion schemes, we're back to just fun f x = f x.
The reason why fun f x = f x has type 'a → 'b is perhaps obvious: The type-inference algorithm assumes that the input type and the output type are 'a and 'b respectively and goes on to conclude the function's only constraint: That f x's input type must be equal to f x's input type, and that f x's output type must be equal to f x's output type, at which point the types 'a and 'b have not been specialized any further.

ML function currying

Could someone please explain the concept of currying to me. I am primarily learning it because we are learning ML in my 'modern programming language' class for a functional language introduction.
In particular you can use this example:
-fun g a = fn b => a+b;
val g = fn: int -> int -> int
-g 2 3;
val it = 5 : int
I'm confused how these parameters are passed or how to even think about it in the first place.
Thank you for any help.
In this case, you make the currying explicit, so it should be easier to understand.
If we read the function definition, it says (paraphrased): "Create a function g, which when given an a returns fn b => a+b."
That is, if we call g 2, we get back the function fn b => 2+b. As such, when we call g 2 3, we actually call (g 2) 3; that is we first get the function stated above back, and then use this function on the value 3, yielding 5.
Currying is simply the concept of making a function in several "stages", each taking an input and producing a new function. SML has syntactic sugar for this, making g equivalent to the following:
fun g a b = a + b;

What does the lambda calculus have to say about return values?

It is by now a well known theorem of the lambda calculus that any function taking two or more arguments can be written through currying as a chain of functions taking one argument:
# Pseudo-code for currying
f(x,y) -> f_curried(x)(y)
This has proven to be extremely powerful not just in studying the behavior of functions but in practical use (Haskell, etc.).
Functions returning values, however, seem to not be discussed. Programmers typically deal with their inability to return more than one value from a function by returning some meta-object (lists in R, structures in C++, etc.). It has always struck me as a bit of a kludge, but a useful one.
For instance:
# R code for "faking" multiple return values
uselessFunc <- function(dat) {
model1 <- lm( y ~ x , data=dat )
return( list( coef=coef(model1), form=formula(model1) ) )
}
Questions
Does the lambda calculus have anything to say about a multiplicity of return values? If so, do any surprising conclusions result?
Similarly, do any languages allow true multiple return values?
According to the Wikipedia page on lambda calculus:
Lambda calculus, also written as λ-calculus, is a formal system for function
definition, function application and recursion
And a function, in the mathematical sense:
Associates one quantity, the argument of the function, also known as the input,
with another quantity, the value of the function, also known as the output
So answering your first question no, lambda calculus (or any other formalism based on mathematical functions) can not have multiple return values.
For your second question, as far as I know, programming languages that implement multiple return values do so by packing multiple results in some kind of data structure (be it a tuple, an array, or even the stack) and then unpacking it later - and that's where the differences lie, as some programming languages make the packing/unpacking part transparent for the programmer (for instance Python uses tuples under the hood) while other languages make the programmer do the job explicitly, for example Java programmers can simulate multiple return values to some extent by packing multiple results in a returned Object array and then extracting and casting the returned result by hand.
A function returns a single value. This is how functions are defined in mathematics. You can return multiple values by packing them into one compound value. But then it is still a single value. I'd call it a vector, because it has components. There are vector functions in mathematics there, so there are also in programming languages. The only difference is the support level from the language itself and does it facilitate it or not.
Nothing prevents you from having multiple functions, each one returning one of the multiple results that you would like to return.
For example, say, you had the following function in python returning a list.
def f(x):
L = []
for i in range(x):
L.append(x * i)
return L
It returns [0, 3, 6] for x=3 and [0, 5, 10, 15, 20] for x=5. Instead, you can totally have
def f_nth_value(x, n):
L = []
for i in range(x):
L.append(x * i)
if n < len(L):
return L[n]
return None
Then you can request any of the outputs for a given input, and get it, or get None, if there aren't enough outputs:
In [11]: f_nth_value(3, 0)
Out[11]: 0
In [12]: f_nth_value(3, 1)
Out[12]: 3
In [13]: f_nth_value(3, 2)
Out[13]: 6
In [14]: f_nth_value(3, 3)
In [15]: f_nth_value(5, 2)
Out[15]: 10
In [16]: f_nth_value(5, 5)
Computational resources may be wasted if you have to do some of the same work, as in this case. Theoretically, it can be avoided by returning another function that holds all the results inside itself.
def f_return_function(x):
L = []
for i in range(x):
L.append(x * i)
holder = lambda n: L[n] if n < len(L) else None
return holder
So now we have
In [26]: result = f_return_function(5)
In [27]: result(3)
Out[27]: 15
In [28]: result(4)
Out[28]: 20
In [29]: result(5)
Traditional untyped lambda calculus is perfectly capable of expressing this idea. (After all, it is Turing complete.) Whenever you want to return a bunch of values, just return a function that can give the n-th value for any n.
In regard to the second question, python allows for such a syntax, if you know exactly, just how many values the function is going to return.
def f(x):
L = []
for i in range(x):
L.append(x * i)
return L
In [39]: a, b, c = f(3)
In [40]: a
Out[40]: 0
In [41]: b
Out[41]: 3
In [42]: c
Out[42]: 6
In [43]: a, b, c = f(2)
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-43-5480fa44be36> in <module>()
----> 1 a, b, c = f(2)
ValueError: need more than 2 values to unpack
In [44]: a, b, c = f(4)
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-44-d2c7a6593838> in <module>()
----> 1 a, b, c = f(4)
ValueError: too many values to unpack
Lastly, here is an example from this Lisp tutorial:
;; in this function, the return result of (+ x x) is not assigned so it is essentially
;; lost; the function body moves on to the next form, (* x x), which is the last form
;; of this function body. So the function call only returns (* 10 10) => 100
* ((lambda (x) (+ x x) (* x x)) 10)
=> 100
;; in this function, we capture the return values of both (+ x x) and (* x x), as the
;; lexical variables SUM and PRODUCT; using VALUES, we can return multiple values from
;; a form instead of just one
* ((lambda (x) (let ((sum (+ x x)) (product (* x x))) (values sum product))) 10)
=> 20 100
I write this as a late response to the accepted answer since it is wrong!
Lambda Calculus does have multiple return values, but it takes a bit to understand what returning multiple values mean.
Lambda Calculus has no inherent definition of a collection of stuff, but it does allow you to invent it using products and church numerals.
pure functional JavaScript will be used for this example.
let's define a product as follows:
const product = a => b => callback => callback(a)(b);
then we can define church_0, and church_1 aka true, false, aka left, right, aka car, cdr, aka first, rest as follows:
const church_0 = a => b => a;
const church_1 = a => b => b;
let's start with making a function that returns two values, 20, and "Hello".
const product = a => b => callback => callback(a)(b);
const church_0 = a => b => a;
const church_1 = a => b => b;
const returns_many = () => product(20)("Hello");
const at_index_zero = returns_many()(church_0);
const at_index_one = returns_many()(church_1);
console.log(at_index_zero);
console.log(at_index_one);
As expected, we got 20 and "Hello".
To return more than 2 values, it gets a bit tricky:
const product = a => b => callback => callback(a)(b);
const church_0 = a => b => a;
const church_1 = a => b => b;
const returns_many = () => product(20)(
product("Hello")(
product("Yes")("No")
)
);
const at_index_zero = returns_many()(church_0);
const at_index_one = returns_many()(church_1)(church_0);
const at_index_two = returns_many()(church_1)(church_1)(church_0);
console.log(at_index_zero);
console.log(at_index_one);
console.log(at_index_two);
As you can see, a function can return an arbitrary number of return values, but to access these values, a you cannot simply use result()[0], result()[1], or result()[2], but you must use functions that filter out the position you want.
This is mindblowingly similar to electrical circuits, in that circuits have no "0", "1", "2", "3", but they do have means to make decisions, and by abstracting away our circuitry with byte(reverse list of 8 inputs), word(reverse list of 16 inputs), in this language, 0 as a byte would be [0, 0, 0, 0, 0, 0, 0, 0] which is equivalent to:
const Byte = a => b => c => d => e => f => g => h => callback =>
callback(a)(b)(c)(d)(e)(f)(g)(h);
const Byte_one = Byte(0)(0)(0)(0)(0)(0)(0)(1); // preserves
const Bit_zero = Byte_one(b7 => b6 => b5 => b4 => b3 => b2 => b1 => b0 => b0);
After inventing a number, we can make an algorithm to, given a byte-indexed array, and a byte representing index we want from this array, it will take care of the boilerplate.
Anyway, what we call arrays is nothing more than the following, expressed in higher level to show the point:
// represent nested list of bits(addresses)
// to nested list of bits(bytes) interpreted as strings.
const MyArray = function(index) {
return (index == 0)
? "0th"
: (index == 1)
? "first"
: "second"
;
};
except it doesnt do 2^32 - 1 if statements, it only does 8 and recursively narrows down the specific element you want. Essentially it acts exactly like a multiplexor(except the "single" signal is actually a fixed number of bits(coproducts, choices) needed to uniquely address elements).
My point is that is Arrays, Maps, Associative Arrays, Lists, Bits, Bytes, Words, are all fundamentally functions, both at circuit level(where we can represent complex universes with nothing but wires and switches), and mathematical level(where everything is ultimately products(sequences, difficult to manage without requiring nesting, eg lists), coproducts(types, sets), and exponentials(free functors(lambdas), forgetful functors)).

Recursive anonymous functions in SML

Is it possible to write recursive anonymous functions in SML? I know I could just use the fun syntax, but I'm curious.
I have written, as an example of what I want:
val fact =
fn n => case n of
0 => 1
| x => x * fact (n - 1)
The anonymous function aren't really anonymous anymore when you bind it to a
variable. And since val rec is just the derived form of fun with no
difference other than appearance, you could just as well have written it using
the fun syntax. Also you can do pattern matching in fn expressions as well
as in case, as cases are derived from fn.
So in all its simpleness you could have written your function as
val rec fact = fn 0 => 1
| x => x * fact (x - 1)
but this is the exact same as the below more readable (in my oppinion)
fun fact 0 = 1
| fact x = x * fact (x - 1)
As far as I think, there is only one reason to use write your code using the
long val rec, and that is because you can easier annotate your code with
comments and forced types. For examples if you have seen Haskell code before and
like the way they type annotate their functions, you could write it something
like this
val rec fact : int -> int =
fn 0 => 1
| x => x * fact (x - 1)
As templatetypedef mentioned, it is possible to do it using a fixed-point
combinator. Such a combinator might look like
fun Y f =
let
exception BlackHole
val r = ref (fn _ => raise BlackHole)
fun a x = !r x
fun ta f = (r := f ; f)
in
ta (f a)
end
And you could then calculate fact 5 with the below code, which uses anonymous
functions to express the faculty function and then binds the result of the
computation to res.
val res =
Y (fn fact =>
fn 0 => 1
| n => n * fact (n - 1)
)
5
The fixed-point code and example computation are courtesy of Morten Brøns-Pedersen.
Updated response to George Kangas' answer:
In languages I know, a recursive function will always get bound to a
name. The convenient and conventional way is provided by keywords like
"define", or "let", or "letrec",...
Trivially true by definition. If the function (recursive or not) wasn't bound to a name it would be anonymous.
The unconventional, more anonymous looking, way is by lambda binding.
I don't see what unconventional there is about anonymous functions, they are used all the time in SML, infact in any functional language. Its even starting to show up in more and more imperative languages as well.
Jesper Reenberg's answer shows lambda binding; the "anonymous"
function gets bound to the names "f" and "fact" by lambdas (called
"fn" in SML).
The anonymous function is in fact anonymous (not "anonymous" -- no quotes), and yes of course it will get bound in the scope of what ever function it is passed onto as an argument. In any other cases the language would be totally useless. The exact same thing happens when calling map (fn x => x) [.....], in this case the anonymous identity function, is still in fact anonymous.
The "normal" definition of an anonymous function (at least according to wikipedia), saying that it must not be bound to an identifier, is a bit weak and ought to include the implicit statement "in the current environment".
This is in fact true for my example, as seen by running it in mlton with the -show-basis argument on an file containing only fun Y ... and the val res ..
val Y: (('a -> 'b) -> 'a -> 'b) -> 'a -> 'b
val res: int32
From this it is seen that none of the anonymous functions are bound in the environment.
A shorter "lambdanonymous" alternative, which requires OCaml launched
by "ocaml -rectypes":
(fun f n -> f f n)
(fun f n -> if n = 0 then 1 else n * (f f (n - 1))
7;; Which produces 7! = 5040.
It seems that you have completely misunderstood the idea of the original question:
Is it possible to write recursive anonymous functions in SML?
And the simple answer is yes. The complex answer is (among others?) an example of this done using a fix point combinator, not a "lambdanonymous" (what ever that is supposed to mean) example done in another language using features not even remotely possible in SML.
All you have to do is put rec after val, as in
val rec fact =
fn n => case n of
0 => 1
| x => x * fact (n - 1)
Wikipedia describes this near the top of the first section.
let fun fact 0 = 1
| fact x = x * fact (x - 1)
in
fact
end
This is a recursive anonymous function. The name 'fact' is only used internally.
Some languages (such as Coq) use 'fix' as the primitive for recursive functions, while some languages (such as SML) use recursive-let as the primitive. These two primitives can encode each other:
fix f => e
:= let rec f = e in f end
let rec f = e ... in ... end
:= let f = fix f => e ... in ... end
In languages I know, a recursive function will always get bound to a name. The convenient and conventional way is provided by keywords like "define", or "let", or "letrec",...
The unconventional, more anonymous looking, way is by lambda binding. Jesper Reenberg's answer shows lambda binding; the "anonymous" function gets bound to the names "f" and "fact" by lambdas (called "fn" in SML).
A shorter "lambdanonymous" alternative, which requires OCaml launched by "ocaml -rectypes":
(fun f n -> f f n)
(fun f n -> if n = 0 then 1 else n * (f f (n - 1))
7;;
Which produces 7! = 5040.

Higher-order type constructors and functors in Ocaml

Can the following polymorphic functions
let id x = x;;
let compose f g x = f (g x);;
let rec fix f = f (fix f);; (*laziness aside*)
be written for types/type constructors or modules/functors? I tried
type 'x id = Id of 'x;;
type 'f 'g 'x compose = Compose of ('f ('g 'x));;
type 'f fix = Fix of ('f (Fix 'f));;
for types but it doesn't work.
Here's a Haskell version for types:
data Id x = Id x
data Compose f g x = Compose (f (g x))
data Fix f = Fix (f (Fix f))
-- examples:
l = Compose [Just 'a'] :: Compose [] Maybe Char
type Natural = Fix Maybe -- natural numbers are fixpoint of Maybe
n = Fix (Just (Fix (Just (Fix Nothing)))) :: Natural -- n is 2
-- up to isomorphism composition of identity and f is f:
iso :: Compose Id f x -> f x
iso (Compose (Id a)) = a
Haskell allows type variables of higher kind. ML dialects, including Caml, allow type variables of kind "*" only. Translated into plain English,
In Haskell, a type variable g can correspond to a "type constructor" like Maybe or IO or lists. So the g x in your Haskell example would be OK (jargon: "well-kinded") if for example g is Maybe and x is Integer.
In ML, a type variable 'g can correspond only to a "ground type" like int or string, never to a type constructor like option or list. It is therefore never correct to try to apply a type variable to another type.
As far as I'm aware, there's no deep reason for this limitation in ML. The most likely explanation is historical contingency. When Milner originally came up with his ideas about polymorphism, he worked with very simple type variables standing only for monotypes of kind *. Early versions of Haskell did the same, and then at some point Mark Jones discovered that inferring the kinds of type variables is actually quite easy. Haskell was quickly revised to allow type variables of higher kind, but ML has never caught up.
The people at INRIA have made a lot of other changes to ML, and I'm a bit surprised they've never made this one. When I'm programming in ML, I might enjoy having higher-kinded type variables. But they aren't there, and I don't know any way to encode the kind of examples you are talking about except by using functors.
You can do something similar in OCaml, using modules in place of types, and functors (higher-order modules) in place of higher-order types. But it looks much uglier and it doesn't have type-inference ability, so you have to manually specify a lot of stuff.
module type Type = sig
type t
end
module Char = struct
type t = char
end
module List (X:Type) = struct
type t = X.t list
end
module Maybe (X:Type) = struct
type t = X.t option
end
(* In the following, I decided to omit the redundant
single constructors "Id of ...", "Compose of ...", since
they don't help in OCaml since we can't use inference *)
module Id (X:Type) = X
module Compose
(F:functor(Z:Type)->Type)
(G:functor(Y:Type)->Type)
(X:Type) = F(G(X))
let l : Compose(List)(Maybe)(Char).t = [Some 'a']
module Example2 (F:functor(Y:Type)->Type) (X:Type) = struct
(* unlike types, "free" module variables are not allowed,
so we have to put it inside another functor in order
to scope F and X *)
let iso (a:Compose(Id)(F)(X).t) : F(X).t = a
end
Well... I'm not an expert of higher-order-types nor Haskell programming.
But this seems to be ok for F# (which is OCaml), could you work with these:
type 'x id = Id of 'x;;
type 'f fix = Fix of ('f fix -> 'f);;
type ('f,'g,'x) compose = Compose of ('f ->'g -> 'x);;
The last one I wrapped to tuple as I didn't come up with anything better...
You can do it but you need to make a bit of a trick:
newtype Fix f = In{out:: f (Fix f)}
You can define Cata afterwards:
Cata :: (Functor f) => (f a -> a) -> Fix f -> a
Cata f = f.(fmap (cata f)).out
That will define a generic catamorphism for all functors, which you can use to build your own stuff. Example:
data ListFix a b = Nil | Cons a b
data List a = Fix (ListFix a)
instance functor (ListFix a) where
fmap f Nil = Nil
fmap f (Cons a lst) = Cons a (f lst)

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