= Fold

This page describes a technique that enables convenient syntax for a
number of language features that are not explicitly supported by
<<StandardML#,Standard ML>>, including: variable number of arguments,
<<OptionalArguments#,optional arguments and labeled arguments>>,
<<ArrayLiteral#,array and vector literals>>,
<<FunctionalRecordUpdate#,functional record update>>,
and (seemingly) dependently typed functions like <<Printf#,printf>> and scanf.

The key idea to _fold_ is to define functions `fold`, `step0`,
and `$` such that the following equation holds.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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 <<EtaExpansion#,eta expand>> the
definition of `f` and `a` to work around the <<ValueRestriction#>>;
we frequently use eta expansion for this purpose without mention.

[source,sml]
----
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`&rsquo;s evaluates to `()`.

[source,sml]
----
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

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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`.

[source,sml]
----
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.

[source,sml]
----
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 `{backtick}` satisfy the following equation.
[source,sml]
----
prod `i1 `i2 ... `im $ = i1 * i2 * ... * im
----

Note that in SML, `{backtick}i1` is two different tokens, `{backtick}` and
`i1`.  We often use `{backtick}` 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).

[source,sml]
----
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.

[source,sml]
----
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:

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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`.

[source,sml]
----
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.

[source,sml]
----
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:

[source,sml]
----
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.

[source,sml]
----
val n = fn z => Fold.fold (0, fn i => i) z
val O = fn z => Fold.step0 (fn i => i * 2) z
val I = fn z => Fold.step0 (fn i => i * 2 + 1) z
----

Here we have one folder, `n`, that can be used with two different
steppers, `O` and `I`.  By using the fold equation, one can
verify the following equations.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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 `{backtick}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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
val IR = fn u => Fold.fold u I R
----

With this definition in place, one can verify the following.

[source,sml]
----
makeSum IR IR $ 1 2.2 3 4.4 = 10.6
----

In general, one can combine steps `s1`, `s2`, ... `sn` as

[source,sml]
----
fn u => Fold.fold u s1 s2 ... sn
----

The following calculation shows why a compound step behaves as the
composition of its constituent steps.

[source,sml]
----
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.

[source,sml]
----
post (w, g) s1 ... sn $ = g (w s1 ... sn $)
----

Here is the definition of `post`.

[source,sml]
----
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)`.

[source,sml]
----
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`.

[source,sml]
----
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.

[source,sml]
----
fun lift0 s (a, f) = fold (fold (a, id) s $, f)
----

Using the definitions, we can prove the following equation.

[source,sml]
----
fold (a, f) (lift0 (step0 h)) = fold (a, f) (step0 h)
----

Here is the proof.

[source,sml]
----
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.

[source,sml]
----
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`.

[source,sml]
----
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.

[source,sml]
----
type ('a, 'b, 'c, 'd) t = ('a, 'b, 'c, 'd) step -> 'd
----

Expanding out the type, we have:

[source,sml]
----
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 <<FirstClassPolymorphism#,first-class polymorphism>>,
we would write the fold type as follows.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
type ('a1, 'a2, 'b, 'c, 'd) step0 =
   ('a1, 'b, 'c, ('a2, 'b, 'c, 'd) t) step
----

Expanding out the type a bit gives:

[source,sml]
----
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 <<FirstClassPolymorphism#,first-class polymorphism>> makes the type
clearer.

[source,sml]
----
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.

[source,sml]
----
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`.

[source,sml]
----
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.

[source,sml]
----
fun step1 (h: 'a11 * 'a12 -> 'a2)
          (a12: 'a12, f: 'b -> 'c)
          (a11: 'a11): ('a2, 'b, 'c, 'd) t =
   fold (h (a11, a12), f)
----

With <<FirstClassPolymorphism#,first-class polymorphism>>, a `step1` function is more clearly
seen as a wrapper around a binary function of type
`'a11 * 'a12 \-> 'a2`.

[source,sml]
----
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`.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
f: 'd fold
a: 'd step
----

Suppose we want to type check

[source,sml]
----
f a a a $: unit
----

As a reminder, the fully parenthesized expression is
[source,sml]
----
((((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

[source,sml]
----
(f a a a) $: unit
----

then we must have

[source,sml]
----
f a a a: unit fold
$: unit step
----

Applying the observation again, we must have

[source,sml]
----
f a a: unit fold fold
a: unit fold step
----

Applying the observation two more times leads to the following type
derivation.

[source,sml]
----
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.

[source,sml]
----
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 <<FirstClassPolymorphism#,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.

[source,sml]
----
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`.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

[source,sml]
----
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.

* <<ArrayLiteral#>>
* <<Fold01N#>>
* <<FunctionalRecordUpdate#>>
* <<NumericLiteral#>>
* <<OptionalArguments#>>
* <<Printf#>>
* <<VariableArityPolymorphism#>>

There are a number of related techniques.  Here are some of them.

* <<StaticSum#>>
* <<TypeIndexedValues#>>
