The key idea to fold is to define functions fold, step0, and $ such that the following equation holds.
fold (a, f) (step0 h1) (step0 h2) ... (step0 hn) $ = f (hn (... (h2 (h1 a))))
The name fold comes because this is like a traditional list fold, where a is the base element, and each step function, step0 hi, corresponds to one element of the list and does one step of the fold. The name $ is chosen to mean end of arguments from its common use in regular-expression syntax.
Unlike the usual list fold in which the same function is used to step over each element in the list, this fold allows the step functions to be different from each other, and even to be of different types. Also unlike the usual list fold, this fold includes a finishing function, f, that is applied to the result of the fold. The presence of the finishing function may seem odd because there is no analogy in list fold. However, the finishing function is essential; without it, there would be no way for the folder to perform an arbitrary computation after processing all the arguments. The examples below will make this clear.
The functions fold, step0, and $ are easy to define.
fun $ (a, f) = f a
fun id x = x
structure Fold =
struct
fun fold (a, f) g = g (a, f)
fun step0 h (a, f) = fold (h a, f)
end
We've placed fold and step0 in the Fold structure but left $ at the toplevel because it is convenient in code to always have $ in scope. We've also defined the identity function, id, at the toplevel since we use it so frequently.
Plugging in the definitions, it is easy to verify the equation from above.
fold (a, f) (step0 h1) (step0 h2) ... (step0 hn) $ = step0 h1 (a, f) (step0 h2) ... (step0 hn) $ = fold (h1 a, f) (step0 h2) ... (step0 hn) $ = step0 h2 (h1 a, f) ... (step0 hn) $ = fold (h2 (h1 a), f) ... (step0 hn) $ ... = fold (hn (... (h2 (h1 a))), f) $ = $ (hn (... (h2 (h1 a))), f) = f (hn (... (h2 (h1 a))))
Example: variable number of arguments
The simplest example of fold is accepting a variable number of (curried) arguments. We'll define a function f and argument a such that all of the following expressions are valid.
f $ f a $ f a a $ f a a a $ f a a a ... a a a $ (* as many a's as we want *)
Off-hand it may appear impossible that all of the above expressions are type correct SML -- how can a function f accept a variable number of curried arguments? What could the type of f be? We'll have more to say later on how type checking works. For now, once we have supplied the definitions below, you can check that the expressions are type correct by feeding them to your favorite SML implementation.
It is simple to define f and a. We define f as a folder whose base element is () and whose finish function does nothing. We define a as the step function that does nothing. The only trickiness is that we must eta expand the definition of f and a to work around the ValueRestriction; we frequently use eta expansion for this purpose without mention.
val base = () fun finish () = () fun step () = () val f = fn z => Fold.fold (base, finish) z val a = fn z => Fold.step0 step z
One can easily apply the fold equation to verify by hand that f applied to any number of a's evaluates to ().
f a ... a $ = finish (step (... (step base))) = finish (step (... ())) ... = finish () = ()
Example: variable-argument sum
Let's look at an example that computes something: a variable-argument function sum and a stepper a such that
sum (a i1) (a i2) ... (a im) $ = i1 + i2 + ... + im
The idea is simple -- the folder starts with a base accumulator of 0 and the stepper adds each element to the accumulator, s, which the folder simply returns at the end.
val sum = fn z => Fold.fold (0, fn s => s) z fun a i = Fold.step0 (fn s => i + s)
Using the fold equation, one can verify the following.
sum (a 1) (a 2) (a 3) $ = 6
Step1
It is sometimes syntactically convenient to omit the parentheses around the steps in a fold. This is easily done by defining a new function, step1, as follows.
structure Fold =
struct
open Fold
fun step1 h (a, f) b = fold (h (b, a), f)
end
From the definition of step1, we have the following equivalence.
fold (a, f) (step1 h) b = step1 h (a, f) b = fold (h (b, a), f)
Using the above equivalence, we can compute the following equation for step1.
fold (a, f) (step1 h1) b1 (step1 h2) b2 ... (step1 hn) bn $ = fold (h1 (b1, a), f) (step1 h2) b2 ... (step1 hn) bn $ = fold (h2 (b2, h1 (b1, a)), f) ... (step1 hn) bn $ = fold (hn (bn, ... (h2 (b2, h1 (b1, a)))), f) $ = f (hn (bn, ... (h2 (b2, h1 (b1, a)))))
Here is an example using step1 to define a variable-argument product function, prod, with a convenient syntax.
val prod = fn z => Fold.fold (1, fn p => p) z val ` = fn z => Fold.step1 (fn (i, p) => i * p) z
The functions prod and ` satisfy the following equation.
prod `i1 `i2 ... `im $ = i1 * i2 * ... * im
Note that in SML, `i1 is two different tokens, ` and i1. We often use ` for an instance of a step1 function because of its syntactic unobtrusiveness and because no space is required to separate it from an alphanumeric token.
Also note that there are no parenthesis around the steps. That is, the following expression is not the same as the above one (in fact, it is not type correct).
prod (`i1) (`i2) ... (`im) $
Example: list literals
SML already has a syntax for list literals, e.g. [w, x, y, z]. However, using fold, we can define our own syntax.
val list = fn z => Fold.fold ([], rev) z val ` = fn z => Fold.step1 (op ::) z
The idea is that the folder starts out with the empty list, the steps accumulate the elements into a list, and then the finishing function reverses the list at the end.
With these definitions one can write a list like:
list `w `x `y `z $
While the example is not practically useful, it does demonstrate the need for the finishing function to be incorporated in fold. Without a finishing function, every use of list would need to be wrapped in rev, as follows.
rev (list `w `x `y `z $)
The finishing function allows us to incorporate the reversal into the definition of list, and to treat list as a truly variable argument function, performing an arbitrary computation after receiving all of its arguments.
See ArrayLiteral for a similar use of fold that provides a syntax for array and vector literals, which are not built in to SML.
Fold right
Just as fold is analogous to a fold left, in which the functions are applied to the accumulator left-to-right, we can define a variant of fold that is analogous to a fold right, in which the functions are applied to the accumulator right-to-left. That is, we can define functions foldr and step0 such that the following equation holds.
foldr (a, f) (step0 h1) (step0 h2) ... (step0 hn) $ = f (h1 (h2 (... (hn a))))
The implementation of fold right is easy, using fold. The idea is for the fold to start with f and for each step to precompose the next hi. Then, the finisher applies the composed function to the base value, a. Here is the code.
structure Foldr =
struct
fun foldr (a, f) = Fold.fold (f, fn g => g a)
fun step0 h = Fold.step0 (fn g => g o h)
end
Verifying the fold-right equation is straightforward, using the fold-left equation.
foldr (a, f) (Foldr.step0 h1) (Foldr.step0 h2) ... (Foldr.step0 hn) $
= fold (f, fn g => g a)
(Fold.step0 (fn g => g o h1))
(Fold.step0 (fn g => g o h2))
...
(Fold.step0 (fn g => g o hn)) $
= (fn g => g a)
((fn g => g o hn) (... ((fn g => g o h2) ((fn g => g o h1) f))))
= (fn g => g a)
((fn g => g o hn) (... ((fn g => g o h2) (f o h1))))
= (fn g => g a) ((fn g => g o hn) (... (f o h1 o h2)))
= (fn g => g a) (f o h1 o h2 o ... o hn)
= (f o h1 o h2 o ... o hn) a
= f (h1 (h2 (... (hn a))))
One can also define the fold-right analogue of step1.
structure Foldr =
struct
open Foldr
fun step1 h = Fold.step1 (fn (b, g) => g o (fn a => h (b, a)))
end
Example: list literals via fold right
Revisiting the list literal example from earlier, we can use fold right to define a syntax for list literals that doesn't do a reversal.
val list = fn z => Foldr.foldr ([], fn l => l) z val ` = fn z => Foldr.step1 (op ::) z
As before, with these definitions, one can write a list like:
list `w `x `y `z $
The difference between the fold-left and fold-right approaches is that the fold-right approach does not have to reverse the list at the end, since it accumulates the elements in the correct order. In practice, MLton will simplify away all of the intermediate function composition, so the the fold-right approach will be more efficient.
Mixing steppers
All of the examples so far have used the same step function throughout a fold. This need not be the case. For example, consider the following.
val n = fn z => Fold.fold (0, fn i => i) z val I = fn z => Fold.step0 (fn i => i * 2) z val O = fn z => Fold.step0 (fn i => i * 2 + 1) z
Here we have one folder, n, that can be used with two different steppers, I and O. By using the fold equation, one can verify the following equations.
n O $ = 0 n I $ = 1 n I O $ = 2 n I O I $ = 5 n I I I O $ = 14
That is, we've defined a syntax for writing binary integer constants.
Not only can one use different instances of step0 in the same fold, one can also intermix uses of step0 and step1. For example, consider the following.
val n = fn z => Fold.fold (0, fn i => i) z val O = fn z => Fold.step0 (fn i => n * 8) z val ` = fn z => Fold.step1 (fn (i, n) => n * 8 + i) z
Using the straightforward generalization of the fold equation to mixed steppers, one can verify the following equations.
n 0 $ = 0 n `3 O $ = 24 n `1 O `7 $ = 71
That is, we've defined a syntax for writing octal integer constants, with a special syntax, O, for the zero digit (admittedly contrived, since one could just write `0 instead of O).
See NumericLiteral for a practical extension of this approach that supports numeric constants in any base and of any type.
(Seemingly) dependent types
A normal list fold always returns the same type no matter what elements are in the list or how long the list is. Variable-argument fold is more powerful, because the result type can vary based both on the arguments that are passed and on their number. This can provide the illusion of dependent types.
For example, consider the following.
val f = fn z => Fold.fold ((), id) z val a = fn z => Fold.step0 (fn () => "hello") z val b = fn z => Fold.step0 (fn () => 13) z val c = fn z => Fold.step0 (fn () => (1, 2)) z
Using the fold equation, one can verify the following equations.
f a $ = "hello": string f b $ = 13: int f c $ = (1, 2): int * int
That is, f returns a value of a different type depending on whether it is applied to argument a, argument b, or argument c.
The following example shows how the type of a fold can depend on the number of arguments.
val grow = fn z => Fold.fold ([], fn l => l) z val a = fn z => Fold.step0 (fn x => [x]) z
Using the fold equation, one can verify the following equations.
grow $ = []: 'a list grow a $ = [[]]: 'a list list grow a a $ = [[[]]]: 'a list list list
Clearly, the result type of a call to the variable argument grow function depends on the number of arguments that are passed.
As a reminder, this is well-typed SML. You can check it out in any implementation.
(Seemingly) dependently-typed functional results
Fold is especially useful when it returns a curried function whose arity depends on the number of arguments. For example, consider the following.
val makeSum = fn z => Fold.fold (id, fn f => f 0) z val I = fn z => Fold.step0 (fn f => fn i => fn x => f (x + i)) z
The makeSum folder constructs a function whose arity depends on the number of I arguments and that adds together all of its arguments. For example, makeSum I $ is of type int -> int and makeSum I I $ is of type int -> int -> int.
One can use the fold equation to verify that the makeSum works correctly. For example, one can easily check by hand the following equations.
makeSum I $ 1 = 1 makeSum I I $ 1 2 = 3 makeSum I I I $ 1 2 3 = 6
Returning a function becomes especially interesting when there are steppers of different types. For example, the following makeSum folder constructs functions that sum integers and reals.
val makeSum = fn z => Foldr.foldr (id, fn f => f 0.0) z val I = fn z => Foldr.step0 (fn f => fn x => fn i => f (x + real i)) z val R = fn z => Foldr.step0 (fn f => fn x: real => fn r => f (x + r)) z
With these definitions, makeSum I R $ is of type int -> real -> real and makeSum R I I $ is of type real -> int -> int -> real. One can use the foldr equation to check the following equations.
makeSum I $ 1 = 1.0 makeSum I R $ 1 2.5 = 3.5 makeSum R I I $ 1.5 2 3 = 6.5
We used foldr instead of fold for this so that the order in which the specifiers I and R appear is the same as the order in which the arguments appear. Had we used fold, things would have been reversed.
An extension of this idea is sufficient to define Printf-like functions in SML.
An idiom for combining steps
It is sometimes useful to combine a number of steps together and name them as a single step. As a simple example, suppose that one often sees an integer follower by a real in the makeSum example above. One can define a new compound step IR as follows.
val IR = fn u => Fold.fold u I R
With this definition in place, one can verify the following.
makeSum IR IR $ 1 2.2 3 4.4 = 10.6
In general, one can combine steps s1, s2, ... sn as
fn u => Fold.fold u s1 s2 ... sn
The following calculation shows why a compound step behaves as the composition of its constituent steps.
fold u (fn u => fold u s1 s2 ... sn) = (fn u => fold u s1 s2 ... sn) u = fold u s1 s2 ... sn
Post composition
Suppose we already have a function defined via fold, w = fold (a, f), and we would like to construct a new fold function that is like w, but applies g to the result produced by w. This is similar to function composition, but we can't just do g o w, because we don't want to use g until w has been applied to all of its arguments and received the end-of-arguments terminator $.
More precisely, we want to define a post-composition function post that satisfies the following equation.
post (w, g) s1 ... sn $ = g (w s1 ... sn $)
Here is the definition of post.
structure Fold =
struct
open Fold
fun post (w, g) s = w (fn (a, h) => s (a, g o h))
end
The following calculations show that post satisfies the desired equation, where w = fold (a, f).
post (w, g) s = w (fn (a, h) => s (a, g o h)) = fold (a, f) (fn (a, h) => s (a, g o h)) = (fn (a, h) => s (a, g o h)) (a, f) = s (a, g o f) = fold (a, g o f) s
Now, suppose si = step0 hi for i from 1 to n.
post (w, g) s1 s2 ... sn $ = fold (a, g o f) s1 s2 ... sn $ = (g o f) (hn (... (h1 a))) = g (f (hn (... (h1 a)))) = g (fold (a, f) s1 ... sn $) = g (w s1 ... sn $)
For a practical example of post composition, see ArrayLiteral.
Lift
We now define a peculiar-looking function, lift0, that is, equationally speaking, equivalent to the identity function on a step function.
fun lift0 s (a, f) = fold (fold (a, id) s $, f)
Using the definitions, we can prove the following equation.
fold (a, f) (lift0 (step0 h)) = fold (a, f) (step0 h)
Here is the proof.
fold (a, f) (lift0 (step0 h)) = lift0 (step0 h) (a, f) = fold (fold (a, id) (step0 h) $, f) = fold (step0 h (a, id) $, f) = fold (fold (h a, id) $, f) = fold ($ (h a, id), f) = fold (id (h a), f) = fold (h a, f) = step0 h (a, f) = fold (a, f) (step0 h)
If lift0 is the identity, then why even define it? The answer lies in the typing of fold expressions, which we have, until now, left unexplained.
Typing
Perhaps the most surprising aspect of fold is that it can be checked by the SML type system. The types involved in fold expressions are complex; fortunately type inference is able to deduce them. Nevertheless, it is instructive to study the types of fold functions and steppers. More importantly, it is essential to understand the typing aspects of fold in order to write down signatures of functions defined using fold and step.
Here is the FOLD signature, and a recapitulation of the entire Fold structure, with additional type annotations.
signature FOLD =
sig
type ('a, 'b, 'c, 'd) step = 'a * ('b -> 'c) -> 'd
type ('a, 'b, 'c, 'd) t = ('a, 'b, 'c, 'd) step -> 'd
type ('a1, 'a2, 'b, 'c, 'd) step0 =
('a1, 'b, 'c, ('a2, 'b, 'c, 'd) t) step
type ('a11, 'a12, 'a2, 'b, 'c, 'd) step1 =
('a12, 'b, 'c, 'a11 -> ('a2, 'b, 'c, 'd) t) step
val fold: 'a * ('b -> 'c) -> ('a, 'b, 'c, 'd) t
val lift0: ('a1, 'a2, 'a2, 'a2, 'a2) step0
-> ('a1, 'a2, 'b, 'c, 'd) step0
val post: ('a, 'b, 'c1, 'd) t * ('c1 -> 'c2)
-> ('a, 'b, 'c2, 'd) t
val step0: ('a1 -> 'a2) -> ('a1, 'a2, 'b, 'c, 'd) step0
val step1: ('a11 * 'a12 -> 'a2)
-> ('a11, 'a12, 'a2, 'b, 'c, 'd) step1
end
structure Fold:> FOLD =
struct
type ('a, 'b, 'c, 'd) step = 'a * ('b -> 'c) -> 'd
type ('a, 'b, 'c, 'd) t = ('a, 'b, 'c, 'd) step -> 'd
type ('a1, 'a2, 'b, 'c, 'd) step0 =
('a1, 'b, 'c, ('a2, 'b, 'c, 'd) t) step
type ('a11, 'a12, 'a2, 'b, 'c, 'd) step1 =
('a12, 'b, 'c, 'a11 -> ('a2, 'b, 'c, 'd) t) step
fun fold (a: 'a, f: 'b -> 'c)
(g: ('a, 'b, 'c, 'd) step): 'd =
g (a, f)
fun step0 (h: 'a1 -> 'a2)
(a1: 'a1, f: 'b -> 'c): ('a2, 'b, 'c, 'd) t =
fold (h a1, f)
fun step1 (h: 'a11 * 'a12 -> 'a2)
(a12: 'a12, f: 'b -> 'c)
(a11: 'a11): ('a2, 'b, 'c, 'd) t =
fold (h (a11, a12), f)
fun lift0 (s: ('a1, 'a2, 'a2, 'a2, 'a2) step0)
(a: 'a1, f: 'b -> 'c): ('a2, 'b, 'c, 'd) t =
fold (fold (a, id) s $, f)
fun post (w: ('a, 'b, 'c1, 'd) t,
g: 'c1 -> 'c2)
(s: ('a, 'b, 'c2, 'd) step): 'd =
w (fn (a, h) => s (a, g o h))
end
That's a lot to swallow, so let's walk through it one step at a time. First, we have the definition of type Fold.step.
type ('a, 'b, 'c, 'd) step = 'a * ('b -> 'c) -> 'd
As a fold proceeds over its arguments, it maintains two things: the accumulator, of type 'a, and the finishing function, of type 'b -> 'c. Each step in the fold is a function that takes those two pieces (i.e. 'a * ('b -> 'c) and does something to them (i.e. produces 'd). The result type of the step is completely left open to be filled in by type inference, as it is an arrow type that is capable of consuming the rest of the arguments to the fold.
A folder, of type Fold.t, is a function that consumes a single step.
type ('a, 'b, 'c, 'd) t = ('a, 'b, 'c, 'd) step -> 'd
Expanding out the type, we have:
type ('a, 'b, 'c, 'd) t = ('a * ('b -> 'c) -> 'd) -> 'd
This shows that the only thing a folder does is to hand its accumulator ('a) and finisher ('b -> 'c) to the next step ('a * ('b -> 'c) -> 'd). If SML had first-class polymorphism, we would write the fold type as follows.
type ('a, 'b, 'c) t = Forall 'd. ('a, 'b, 'c, 'd) step -> 'd
This type definition shows that a folder had nothing to do with the rest of the fold, it only deals with the next step.
We now can understand the type of fold, which takes the initial value of the accumulator and the finishing function, and constructs a folder, i.e. a function awaiting the next step.
val fold: 'a * ('b -> 'c) -> ('a, 'b, 'c, 'd) t
fun fold (a: 'a, f: 'b -> 'c)
(g: ('a, 'b, 'c, 'd) step): 'd =
g (a, f)
Continuing on, we have the type of step functions.
type ('a1, 'a2, 'b, 'c, 'd) step0 =
('a1, 'b, 'c, ('a2, 'b, 'c, 'd) t) step
Expanding out the type a bit gives:
type ('a1, 'a2, 'b, 'c, 'd) step0 =
'a1 * ('b -> 'c) -> ('a2, 'b, 'c, 'd) t
So, a step function takes the accumulator ('a1) and finishing function ('b -> 'c), which will be passed to it by the previous folder, and transforms them to a new folder. This new folder has a new accumulator ('a2) and the same finishing function.
Again, imagining that SML had first-class polymorphism makes the type clearer.
type ('a1, 'a2) step0 =
Forall ('b, 'c). ('a1, 'b, 'c, ('a2, 'b, 'c) t) step
Thus, in essence, a step0 function is a wrapper around a function of type 'a1 -> 'a2, which is exactly what the definition of step0 does.
val step0: ('a1 -> 'a2) -> ('a1, 'a2, 'b, 'c, 'd) step0
fun step0 (h: 'a1 -> 'a2)
(a1: 'a1, f: 'b -> 'c): ('a2, 'b, 'c, 'd) t =
fold (h a1, f)
It is not much beyond step0 to understand step1.
type ('a11, 'a12, 'a2, 'b, 'c, 'd) step1 =
('a12, 'b, 'c, 'a11 -> ('a2, 'b, 'c, 'd) t) step
A step1 function takes the accumulator ('a12) and finisher ('b -> 'c) passed to it by the previous folder and transforms them into a function that consumes the next argument ('a11) and produces a folder that will continue the fold with a new accumulator ('a2) and the same finisher.
fun step1 (h: 'a11 * 'a12 -> 'a2)
(a12: 'a12, f: 'b -> 'c)
(a11: 'a11): ('a2, 'b, 'c, 'd) t =
fold (h (a11, a12), f)
With first-class polymorphism, a step1 function is more clearly seen as a wrapper around a binary function of type 'a11 * 'a12 -> 'a2.
type ('a11, 'a12, 'a2) step1 =
Forall ('b, 'c). ('a12, 'b, 'c, 'a11 -> ('a2, 'b, 'c) t) step
The type of post is clear: it takes a folder with a finishing function that produces type 'c1, and a function of type 'c1 -> 'c2 to postcompose onto the folder. It returns a new folder with a finishing function that produces type 'c2.
val post: ('a, 'b, 'c1, 'd) t * ('c1 -> 'c2)
-> ('a, 'b, 'c2, 'd) t
fun post (w: ('a, 'b, 'c1, 'd) t,
g: 'c1 -> 'c2)
(s: ('a, 'b, 'c2, 'd) step): 'd =
w (fn (a, h) => s (a, g o h))
We will return to lift0 after an example.
An example typing
Let's type check our simplest example, a variable-argument fold. Recall that we have a folder f and a stepper a defined as follows.
val f = fn z => Fold.fold ((), fn () => ()) z val a = fn z => Fold.step0 (fn () => ()) z
Since the accumulator and finisher are uninteresting, we'll use some abbreviations to simplify things.
type 'd step = (unit, unit, unit, 'd) Fold.step type 'd fold = 'd step -> 'd
With these abbreviations, f and a have the following polymorphic types.
f: 'd fold a: 'd step
Suppose we want to type check
f a a a $: unitAs a reminder, the fully parenthesized expression is
((((f a) a) a) a) $The observation that we will use repeatedly is that for any type z, if f: z fold and s: z step, then f s: z. So, if we want
(f a a a) $: unitthen we must have
f a a a: unit fold $: unit stepApplying the observation again, we must have
f a a: unit fold fold a: unit fold step
Applying the observation two more times leads to the following type derivation.
f: unit fold fold fold fold a: unit fold fold fold step f a: unit fold fold fold a: unit fold fold step f a a: unit fold fold a: unit fold step f a a a: unit fold $: unit step f a a a $: unit
So, each application is a fold that consumes the next step, producing a fold of one smaller type.
One can expand some of the type definitions in f to see that it is indeed a function that takes four curried arguments, each one a step function.
f: unit fold fold fold step -> unit fold fold step -> unit fold step -> unit step -> unit
This example shows why we must eta expand uses of fold and step0 to work around the value restriction and make folders and steppers polymorphic. The type of a fold function like f depends on the number of arguments, and so will vary from use to use. Similarly, each occurrence of an argument like a has a different type, depending on the number of remaining arguments.
This example also shows that the type of a folder, when fully expanded, is exponential in the number of arguments: there are as many nested occurrences of the fold type constructor as there are arguments, and each occurrence duplicates its type argument. One can observe this exponential behavior in a type checker that doesn't share enough of the representation of types (e.g. one that represents types as trees rather than directed acyclic graphs).
Generalizing this type derivation to uses of fold where the accumulator and finisher are more interesting is straightforward. One simply includes the type of the accumulator, which may change, for each step, and the type of the finisher, which doesn't change from step to step.
Typing lift
The lack of first-class polymorphism in SML causes problems if one wants to use a step in a first-class way. Consider the following double function, which takes a step, s, and produces a composite step that does s twice.
fun double s = fn u => Fold.fold u s s
The definition of double is not type correct. The problem is that the type of a step depends on the number of remaining arguments but that the parameter s is not polymorphic, and so can not be used in two different positions.
Fortunately, we can define a function, lift0, that takes a monotyped step function and lifts it into a polymorphic step function. This is apparent in the type of lift0.
val lift0: ('a1, 'a2, 'a2, 'a2, 'a2) step0
-> ('a1, 'a2, 'b, 'c, 'd) step0
fun lift0 (s: ('a1, 'a2, 'a2, 'a2, 'a2) step0)
(a: 'a1, f: 'b -> 'c): ('a2, 'b, 'c, 'd) t =
fold (fold (a, id) s $, f)
The following definition of double uses lift0, appropriately eta wrapped, to fix the problem.
fun double s =
let
val s = fn z => Fold.lift0 s z
in
fn u => Fold.fold u s s
end
With that definition of double in place, we can use it as in the following example.
val f = fn z => Fold.fold ((), fn () => ()) z val a = fn z => Fold.step0 (fn () => ()) z val a2 = fn z => double a z val () = f a a2 a a2 $
Of course, we must eta wrap the call double in order to use its result, which is a step function, polymorphically.
Hiding the type of the accumulator
For clarity and to avoid mistakes, it can be useful to hide the type of the accumulator in a fold. Reworking the simple variable-argument example to do this leads to the following.
structure S:>
sig
type ac
val f: (ac, ac, unit, 'd) Fold.t
val s: (ac, ac, 'b, 'c, 'd) Fold.step0
end =
struct
type ac = unit
val f = fn z => Fold.fold ((), fn () => ()) z
val s = fn z => Fold.step0 (fn () => ()) z
end
The idea is to name the accumulator type and use opaque signature matching to make it abstract. This can prevent improper manipulation of the accumulator by client code and ensure invariants that the folder and stepper would like to maintain.
For a practical example of this technique, see ArrayLiteral.
Also see
Fold has a number of practical applications. Here are some of them.