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# Continuations
Last week we saw how to turn a list zipper into a continuation-based
list processor. The function computed something we called "the task",
which was a simplified langauge involving control operators.
abSdS ~~> ababdS ~~> ababdababd
The task is to process the list from left to right, and at each "S",
double the list so far. Here, "S" is a kind of control operator, and
captures the entire previous computation. We also considered a
variant in which '#' delimited the portion of the list to be copied:
ab#deSfg ~~> abededfg
In this variant, "S" and "#" correspond to `shift` and `reset`, which
provide access to delimited continuations.
The expository logic of starting with this simplified task is the
notion that as lists are to trees, so is this task to full-blown
continuations. So to the extent that, say, list zippers are easier to
grasp than tree zippers, the task is easier to grasp than full
continuations.
We then presented CPS transforms, and demonstrated how they provide
an order-independent analysis of order of evaluation.
In order to continue to explore continuations, we will proceed in the
followin fashion.
## The continuation monad
Let's take a look at some of our favorite monads from the point of
view of types. Here, `==>` is the Kleisli arrow.
Reader monad: types: Mα ==> β -> α
⇧: \ae.a : α -> Mα
compose: \fghe.f(ghe)e : (Q->MR)->(P->MQ)->(P->MR)
gloss: copy environment and distribute it to f and g
State monad: types: α ==> β -> (α x β)
⇧: \ae.(a,e) : α -> Mα
compose: \fghe.let (x,s) = ghe in fxs
thread the state through g, then through f
List monad: types: α ==> [α]
⇧: \a.[a] : α -> Mα
compose: \fgh.concat(map f (gh))
gloss: compose f and g pointwise
Maybe monad: types: α ==> Nothing | Just α
⇧: \a. Just a
compose: \fgh.
case gh of Nothing -> Nothing
| Just x -> case fx of Nothing -> Nothing
| Just y -> Just y
gloss: strong Kline
Now we need a type for a continuation. A continuized term is one that
expects its continuation as an argument. The continuation of a term
is a function from the normal value of that term to a result. So if
the term has type continuized term has type α, the continuized version
has type (α -> ρ) -> ρ:
Continuation monad: types: Mα => (α -> ρ) -> ρ
⇧: \ak.ka
compose: \fghk.f(\f'.g h (\g'.f(g'h)
gloss: first give the continuation to f, then build
a continuation out of the result to give to
The first thing we should do is demonstrate that this monad is
suitable for accomplishing the task.
We lift the computation `("a" ++ ("b" ++ ("c" ++ "d")))` into
the monad as follows:
t1 = (map1 ((++) "a") (map1 ((++) "b") (map1 ((++) "c") (mid "d"))))
Here, `(++) "a"` is a function of type [Char] -> [Char] that prepends
the string "a", so `map1 ((++) "a")` takes a string continuation k and
returns a new string continuation that takes a string s returns "a" ++ k(s).
So `t1 (\k->k) == "abcd"`.
We can now define a shift operator to perform the work of "S":
shift u k = u(\s.k(ks))
Shift takes two arguments: a string continuation u of type (String -> String) -> String,
and a string continuation k of type String -> String. Since u is the
result returned by the argument to shift, it represents the tail of
the list after the shift operator. Then k is the continuation of the
expression headed by `shift`. So in order to execute the task, shift
needs to invoke k twice.
Note that the shift operator constructs a new continuation by
composing its second argument with itself (i.e., the new doubled
continuation is \s.k(ks)). Once it has constructed this
new continuation, it delivers it as an argument to the remaining part
of the computation!
(map1 ((++) "a") (map1 ((++) "b") (shift (mid "d")))) (\k->k) == "ababd"
Let's just make sure that we have the left-to-right evaluation we were
hoping for by evaluating "abSdeSf":
t6 = map1 ((++) "a")
(map1 ((++) "b")
(shift
(map1 ((++) "d")
(map1 ((++) "e")
(shift (mid "f"))))))
t6 (\k->k) == "ababdeababdef"
As expected.
In order to add a reset operator #, we can have
# u k = k(u(\k.k))
ab#deSf ~~> abdedef
Note that the lifting of the original unmonadized computation treated
prepending "a" as a one-place operation. If we decompose this
operation into a two-place operation of appending combined with a
string "a", an interesting thing happens.
map2 f u v k = u(\u' -> v (\v' -> k(f u' v')))
shift k = k (k "")
t2 = map2 (++) (mid "a")
(map2 (++) (mid "b")
(map2 (++) shift
(map2 (++) (mid "d")
(mid []))))
t2 (\k->k) == "ababdd"
First, we need map2 instead of map1. Second, the type of the shift
operator will be a string continuation, rather than a function from
string continuations to string continuations.
But here's the interesting part: from the point of view of the shift
operator, the continuation that it will be fed will be the entire rest
of the computation. This includes not only what has come before, but
what will come after it as well. That means when the continuation is
doubled (self-composed), the result duplicates not only the prefix
(ab ~~> abab), but also the suffix (d ~~> dd). In some sense, then,
continuations allow functions to see into the future!
What do you expect will be the result of executing "aScSe" under the
second perspective?
This depends on which S is executed first. Assume the first S is
executed first. What will the value of its continuation k be?
It will be a function that maps a string s to the result of computing
ascSe, which will be ascascee. So k(k "") will be k(acacee), which
will result in a(acacee)ca(acacee)cee (the parenthesese are added to
show stages in the construction of the final result).
Note that this is a different result than assuming that what execution
does is choose an S, and double its context, treating all other shift
operators as dumb letters, then choosing a remaining S to execute. If
that was our interpreation of the task, then no string with more than
one S would ever terminate (on the bi-directional continuation
perspective).
## Viewing Montague's PTQ as CPS
## How continuations can simulate other monads
## Delimited versus undelimited continuations
## Natural language requries delimited continuations