MP 4 – Continuation-Passing StyleCS 421 – Fall 2012Revis.docx

MP 4 – Continuation-Passing Style
CS 421 – Fall 2012
Revision 1.0
Assigned September 17, 2013
Due September 24, 2013 21:59
Extension 48 hours (20% penalty)
1 Change Log
1.0 Initial Release.
2 Objectives and Background
The purpose of this MP is to help the student learn the basics of
continuation-passing style, or CPS, and CPS transfor-
mation. Next week, you will be using your knowledge learned
from this MP to construct a general-purpose algorithm
for transforming code in direct style into continuation-passing
style.
3 Instructions
The problems below are all similar to the problems in MP2 and
MP3. The difference is that you must implement each
of these function in continuation-passing style. In some cases,
you must first write a function in direct style (according
to the problem specification), then transform the function
definition into continuation-passing style.
The problems below have sample executions that suggest how to
write answers. Students have to use the same
function name, but the name of the parameters that follow the
function name need not be duplicated. That is, the

students are free to choose different names for the arguments to
the functions from the ones given in the example
execution. We also will use let rec to begin the definition of
some of the functions that are allowed to use recursion.
You are not required to start your code with let rec. Similarly, if
you are not prohibited from using explicit
recursion in a given problem, you may change any function
definition from starting with just let to starting with let
rec.
For all these problems, you are allowed to write your own
auxiliary functions, either internally to the function
being defined or externally as separate functions. All such
helper functions must satisfy any coding restrictions (such
as being in tail recursive form, or not using explicit recursion)
as the main function being defined for the problem must
satisfy.
Here is a list of the strict requirements for the assignment.
• The function name must be the same as the one provided.
• The type of parameters must be the same as the parameters
shown in sample execution.
• Students must comply with any special restrictions for each
problem. For several of the problems, you will be
required to write a function in direct style, possibly with some
restrictions, as you would have in MP2 or MP3,
and then transform the code you wrote in continuation-passing
style.
1

4 Problems
These exercises are designed to give you a feel for
continuation-passing style. A function that is written in
continuation-
passing style does not return once it has finished computing.
Instead, it calls another function (the continuation) with
the result of the computation. Here is a small example:
# let report x =
print_string "Result: ";
print_int x;
print_newline();;
val report : int -> unit = <fun>
# let inck i k = k (i+1)
val inck : int -> (int -> ’a) -> ’a = <fun>
The inck function takes an integer and a continuation. After
adding 1 to the integer, it passes the result to its
continuation.
# inck 3 report;;
Result: 4
- : unit = ()
# inck 3 inck report;;
Result: 5
- : unit = ()
In line 1, inck increments 3 to be 4, and then passes the 4 to
report. In line 4, the first inck adds 1 to 3, and
passes the resulting 4 to the second inck, which then adds 1 to
4, and passes the resulting 5 to report.
4.1 Transforming Primitive Operations
Primitive operations are “transformed” into functions that take

the arguments of the original operation and a continu-
ation, and apply the continuation to the result of applying the
primitive operation on its arguments.
1. (10 pts) Write the following low-level functions in
continuation-passing style. A description of what each function
should do follows:
• addk adds two integers;
• subk subtracts the second integer from the first;
• mulk multiplies two integers;
• posk determines if the argument is strictly positive;
• float addk adds two floats;
• float divk divides the first float by the second float;
• catk concatenates two strings;
• consk created a new list by adding an element at the front of a
list;
• geqk determines if the first argument is greater than or equal
to the second argument; and
• eqk determines if the two arguments are equal.
# let addk n m k = ...;;
val addk : int -> int -> (int -> ’a) -> ’a = <fun>
# let subk n m k = ...;;
val subk : int -> int -> (int -> ’a) -> ’a = <fun>
2
# let mulk n m k = ...;;
val mulk : int -> int -> (int -> ’a) -> ’a = <fun>
# let posk x k = ...;;
val posk : int -> (bool -> ’a) -> ’a = <fun>
# let float_addk a b k = ...;;
val float_addk : float -> float -> (float -> ’a) -> ’a = <fun>

# let float_divk a b k = ...;;
val float_divk : float -> float -> (float -> ’a) -> ’a = <fun>
# let catk str1 str2 k = ...;;
val catk : string -> string -> (string -> ’a) -> ’a = <fun>
# let consk e l k = ...;;
val consk : ’a -> ’a list -> (’a list -> ’b) -> ’b = <fun>
# let eqk x y k = ...;;
val eqk : ’a -> ’a -> (bool -> ’b) -> ’b = <fun>
# let geqk x y k = ...;;
val geqk : ’a -> ’a -> (bool -> ’b) -> ’b = <fun>
# addk 1 1 report;;
Result: 2
- : unit = ()
# catk "hello " "world" (fun x -> x);;
- : string = "hello world"
# float_addk 1.0 1.0
(fun x -> float_divk x 2.0
(fun y -> (print_string "Result: ";print_float y;
print_newline())));;
Result: 1.
- : unit = ()
# geqk 2 1 (fun b -> (report (if b then 1 else 0)));;
Result: 1
- : unit = ()
4.2 Nesting Continuations
# let add3k a b c k =
addk a b (fun ab -> addk ab c k);;
val add3k : int -> int -> int -> (int -> ’a) -> ’a = <fun>
# add3k 1 2 3 report;;
Result: 6
- : unit = ()

We needed to add three numbers together, but addk itself only
adds two numbers. On line 2, we give the first call
to addk a function that saves the sum of a and b in the variable
ab. Then this function adds ab to c and passes its
result to the continuation k.
2. (5 pts) Using addk and mulk as helper functions, write a
function polyk, which takes on integer argument x
and “returns” x3 + x + 1. You may only use the addk and mulk
operators to do the arithmetic. The order of
evaluation of operations must be as follows: first compute x3,
then compute x + 1, and then compute x3 + x + 1.
# let poly x k = ...;;
val poly : int -> (int -> ’a) -> ’a = <fun>
# poly 2 report;;
Result: 11
- : unit = ()
3
3. (5 pts) Write a function composek, which takes as the first
two arguments two functions f and g and “returns”
g ◦ f (the composition of f with g). A function h is equal with g
◦ f if and only if h(x) = g(f(x)) for all elements
in the domain of f. You must assume that the functions f and g
are given in the continuation-passing style.
# let compose f g k = ...;;
val composek :
(’a -> (’b -> ’c) -> ’d) ->
(’b -> ’e -> ’c) -> ((’a -> ’e -> ’d) -> ’f) -> ’f = <fun>

# composek inck inck (fun h -> h 1 report);;
Result: 3
- : unit = ()
4.3 Transforming Recursive Functions
How do we write recursive programs in CPS? Consider the
following recursive function:
# let rec factorial n =
if n = 0 then 1 else n * factorial (n - 1);;
val factorial : int -> int = <fun>
# factorial 5;;
- : int = 120
We can rewrite this making each step of computation explicit as
follows:
# let rec factoriale n =
let b = n = 0 in
if b then 1
else let s = n - 1 in
let m = factoriale s in
n * m;;
val factoriale : int -> int = <fun>
# factoriale 5;;
- : int = 120
To put the function into full CPS, we must make factorial take
an additional argument, a continuation, to which the
result of the factorial function should be passed. When the
recursive call is made to factorial, instead of it returning a
result to build the next higher factorial, it needs to take a

continuation for building that next value from its result. In
addition, each intermediate computation must be converted so
that it also takes a continuation. Thus the code becomes:
# let rec factorialk n k =
eqk n 0
(fun b -> if b then k 1
else subk n 1
(fun s -> factorialk s
(fun m -> timesk n m k)));;
# factorialk 5 report;;
Result: 120
- : unit = ()
Notice that to make a recursive call, we needed to build an
intermediate continuation capturing all the work that must
be done after the recursive call returns and before we can return
the final result. If m is the result of the recursive call
in direct style (without continuations), then we need to build a
continuation to:
4
• take the recursive value: m
• build it to the final result: n * m
• pass it to the final continuation k
Notice that this is an extension of the ”nested continuation”
method.
In Problems 4 through 6 you are asked to first write a function

in direct style and then transform the code into
continuation-passing style. When writing functions in
continuation-passing style, all uses of functions need to take a
continuation as an argument. For example, if a problem asks
you to write a function partition, then you should
define partition in direct style and partitionk in continuation-
passing style. All uses of primitive operations
(e.g. +, -, *, >=, =) should use the corresponding functions
defined in Problem 1. If you need to make use of
primitive operations not covered in Problem 1, you should
include a definition of the corresponding version that takes
a continuation as an additional argument, as in Problem 1. In
Problem 5 and 6, there must be no use of list library
functions.
4. (6 pts total)
a. (2 pts) Write a function inverse square series, which takes an
integer n, and computes the (partial)
series 1 + 1
4
+ 1
9
+ . . . + 1
n2
and returns the result. If n ≤ 0, then return 0.
# let rec inverse_square_series n = ...;;
val inverse_square_series : int -> float = <fun>
# inverse_square_series 10;;
- : float = 2.92896825396825378

b. (4 pts) Write the function inverse square series which is the
CPS transformation of the code you
wrote in part a.
# let rec inverse_square_seriesk n k = ...;;
val inverse_square_seriesk : int -> (float -> ’a) -> ’a = <fun>
# inverse_square_seriesk 10 (fun x -> x);;
- : float = 2.92896825396825378
5. (8 pts total)
a. (2 pts) Write a function rev map which takes a function f (of
type ’a -> ’b) and a list l (of type ’a
list). If the list l has the form [a1; a2; . . . ; an], then rev map
applies f on an, then on an−1, then . . . ,
then on a1 and then builds the list [f(a1); . . . ; f(an)] with the
results returned by f. The function f may
have side-effects (e.g. report). There must be no use of list
library functions.
# let rec rev_map f l = ...;;
val rev_map : (’a -> ’b) -> ’a list -> ’b list = <fun>
# rev_map (fun x -> print_int x; x + 1) [1;2;3;4;5];;
54321- : int list = [2; 3; 4; 5; 6]
b. (6 pts) Write the function rev mapk that is the CPS
transformation of the code you wrote in part a. You
must assume that the function f is also transformed in
continuation-passing style, that is, the type of f is not
’a -> ’b, but ’a -> (’b -> ’c) -> ’c.
# let rec rev_mapk f l k = ...;;
val rev_mapk :
(’a -> (’b -> ’c) -> ’c) -> ’a list -> (’b list -> ’c) -> ’c = <fun>
# let print_intk i k = k (print_int i);;
val print_intk : int -> (unit -> ’a) -> ’a = <fun>
# rev_mapk (fun x -> fun k -> print_intk x (fun t -> inck x k))

[1;2;3;4;5] (fun x -> x);;
54321- : int list = [2; 3; 4; 5; 6]
5
6. (8 pts total)
a. (2 pts) Write a function partition which takes a list l (of type
’a list), and a predicate p (of type
’a -> bool), and returns a pair of lists (l1, l2) where l1 contains
all the elements in l satisfying p, and l2
contains all the elements in l not satisfying p. The order of the
elements in l1 and l2 is the order in l. There
must be no use of list library functions.
# let rec partition l p = ...;;
val partition : ’a list -> (’a -> bool) -> ’a list * ’a list = <fun>
# partition [1;2;3;4;5] (fun x -> x >= 3);;
- : int list * int list = ([3; 4; 5], [1; 2])
b. (6 pts) Write a function partitionk which is the CPS
transformation of the code you wrote in part a. You
must assume that the predicate p is also transformed in
continuation-passing style, that is, its type is not ’a
-> bool, but ’a -> (bool -> ’b) -> ’b.
# let rec let rec partitionk l p k = ...;;
val partitionk :
’a list -> (’a -> (bool -> ’b) -> ’b) -> (’a list * ’a list -> ’b)
-> ’b = <fun>
# partitionk [1;2;3;4;5] (fun x -> fun k -> geqk x 3 k) (fun x ->
x);;

- : int list * int list = ([3; 4; 5], [1; 2])
4.4 Using Continuations to Alter Control Flow
As we have seen in the previous sections, continuations allow
us a way of explicitly stating the order of events, and
in particular, what happens next. We can use this abilty to
increase our flexibility over the control of the flow of
execution (referred to as control flow). If we build and keep at
our access several different continuations, then we have
the ability to choose among them which one to use in moving
forward, thereby altering our flow of execution. You are
all familiar with using an if-then-else as a control flow
construct to enable the program to dynamically choose between
two different execution paths going forward.
Another useful control flow construct is that of raising and
handling exceptions. In class, we gave an example of
how we can use continuations to abandon the current execution
path and roll back to an earlier point to continue with
a different path of execution from that point. This method
involves keeping track of two continuations at the same
time: a primary one that handles “normal” control flow, and one
that remembers the point to roll back to when an
exceptional case turns up. As in regular continuation-passing
style, the primary continuation should be continuously
updated; however, the exception continuation remains the same.
The exception continuation is then passed the control
flow (by being called) when an exceptional state comes up, and
the primary continuation is used otherwise.
7. (8 pts) Write a function findk which takes a list l, a predicate
p, a normal continuation (of type ’a -> ’b),
and an exception continuation (of type unit -> ’b), and searches
for the first element in l satisfying p. If
there is such an element, it calls the normal continuation with
the said element; otherwise, it calls the exception

continuation with the unit (“()”). Your definition must be in
continuation-passing style, and must follow the same
restrictions about calling primitives as the previous section’s
problems. (For this problem, you will receive no
points for the direct style definition of find, though writing it
may be helpful when converting it to continuation-
passing style.)
# let rec findk l p normalk exceptionk = ...;;
val findk :
’a list -> (’a -> (bool -> ’b) -> ’b) -> (’a -> ’b) -> (unit -> ’b)
-> ’b = <fun>
# findk [1;2;3;4;5] (fun x -> fun k -> eqk x 3 k)
(fun x -> x) (fun x -> print_string "element not found"; -1);;
6
- : int = 3
# findk [1;2;3;4;5] (fun x -> fun k -> eqk x 6 k)
(fun x -> x) (fun x -> print_string "element not found"; -1);;
element not found- : int = -1
4.5 Extra Credit
8. (8 pts)
Write the function appk which takes a list l of functions in
continuation-passing style (of type ’a -> (’a ->
’b) -> ’b), an initial value x (of type ’a) and a continuation k. If
the list l is of the form [f1; . . . ; fn], then
appk evaluates f1 (f2 (. . . (fn x) . . .)) and passes the result to
k. Intuitively, it evaluates fn on x, then fn−1 on

the result, then fn−2 on the second result, and so on. Your
definition must be in continuation-passing style.
# let rec appk l x k = ...;
val appk : (’a -> (’a -> ’b) -> ’b) list -> ’a -> (’a -> ’b) -> ’b =
<fun>
# appk [inck;inck;inck] 0 (fun x -> x);;
- : int = 3
7
HW 4 – CSP Transformation; Working with
Mathematical Specifications
CS 421 – Fall 2013
Revision 1.0
Assigned Wednesday, September 18, 2013
Due Wednesday, September 25, 2013, 19:59pm
Extension 48 hours (20% penalty)
1 Change Log
1.0 Initial Release.
2 Objectives and Background
The purpose of this HW is to help your understanding of:
• The basic CSP transformation algorithm
• How to use a formal mathematically recursive definition
3 Turn-In Procedure
Using your favorite tool(s), you should put your solution in a

file named hw4.pdf (the same name as this file has
on the website). If you have problems generating a pdf, please
seek help from the course staff. Your answers to the
following questions are to be submitted electronically via the
handin script as though an MP. This assignment is
named hw4.
4 Background and Definitions
Throughout this HW, we will be working with a (very) simple
functional language. It is a fragment of MicorML (a
frgament of SML, an OCaml-like language), and the seed of the
language that we will be using in MPs throughout
the rest of this semester. Using a mix of MicroML concrete
syntax (expression constructs as you would type them
in MicroML’s top-level loop) and recursive mathematical
functions, we will describe below the algorithm for CSP
transformation for this fragment. You should compare this
formal definition with the description given on examples in
class.
The language fragment we will work with in this assignment is
given by the following Context Free Grammar:
e → c | v | e⊕e
| if e then e else e
| fn v => ; | e e
The symbol e ranges recursively over all expressions in
MicroML, c ranges over constants, v ranges over program
variables, and ⊕ ranges over infixed binary primitive
operations. The MicroML construct fn v=> e corresponds the
OCaml’s construct fun v-> e. This language will expand over
the course of the semester.
1

Mathematically we represent CPS transformation by the
functions [[e]]κ, which calculates the CPS form of an
expression e when passed the continuation κ. The symbol κ does
not represent a programming language variable, but
rather a complex expression describing the current continuation
for e.
The defining equations of this function are given below. Recall
that when transforming a function into CPS, it is
necessary to expand the arguments to the function to include
one that is for passing the continuation to it. We will use
κ to represent a continuation that has already been calculated or
given to us, and k, ki, k′ etc as the name of variables
that can be assigned continuations. We will use v, vi, v′ etc for
the ordinary variables in our program.
By v being fresh for an expression e, we mean that v needs to be
some variable that is NOT in e. In MP5, you will
implement a function for selecting one, but here you are free to
choose a name as you please, subject to being different
from all the other names that have already been used.
• The CPS transformation of a variable or constant expression
just applies the continuation to the variable or
constant, since during execution, when this point in the code is
reached, both variables and constants are already
fully evaluated (except for being looked up).
[[v]]κ = κ v
[[c]]κ = κ c
Example: [[x]](FUN y -> report y) = (FUN y -> report y) x This
may be read as “load reg-

ister y with the value for x, then do a procedure call to report”.
• The CPS transformation for a binary operator mirrors its
evaluation order. It first evaluates its first argument
then its second before evaluating the binary operator applied to
those two values. We create a new continuation
that takes the result of the first argument, e1, binds it to v1 then
evaluates the second argument, e2, and binds
that result to v2. As a last step it applies the current
continuation to the result of v1 ⊕v2. This is formalized in
the following rule:
[[e1 ⊕e2]]κ = [[e1]]FUN v1 -> [[e2]]FUN v2 -> κ (v1 ⊕v2)
Where
v1 is fresh for e1, e2, and κ
v2 is fresh for e1, e2, κ, and v1
Example: [[x + 1]](FUN w -> report w)
= [[x]](FUN y -> [[1]](FUN z -> (FUN w -> report w) (y + z)))
= [[x]](FUN y -> ((FUN z -> (FUN w -> report w) (y + z)) 1))
= (FUN y -> ((FUN z -> (FUN w -> report w) (y + z)) 1)) x
• Each CPS transformation should make explicit the order of
evaluation of each subexpression. For if-then-else
expressions, the first thing to be done is to evaluate the boolean
guard. The resulting boolean value needs to
be passed to an if-then-else that will choose a branch. When the
boolean value is true, we need to evaluate
the transformed then-branch, which will pass its value to the
final continuation for the if-then-else expression.
Similarly, when the boolean value is false we need to evaluate
the transformed else-branch, which will pass its
value to the final continuation for the if-then-else expression.
To accomplish this, we recursively CPS-transform
the boolean guard e1 with the continuation with a formal

parameter v that is fresh for the then branch e2, the
else branch e3 and the final continuation κ, where, based on the
value of v, the continuation chooses either
the CPS-transform of e2 with the original continuation κ, or the
CPS-transform of e3, again with the original
continuation κ.
[[if e1 then e2 else e3]]κ = [[e1]]FUN v -> IF v THEN [[e2]]κ
ELSE [[e3]]κ
Where v is fresh for e2, e3, and κ
2
With FUN v -> IF v THEN [[e2]]κ ELSE [[e3]]κ we are creating
a new continutation from our old. This
is not a function at the level of expressions, but rather at the
level of continuations, hence the use of a different,
albeit related, syntax.
Example:
[[if x > 0 then 2 else 3]](FUN w -> report w)
= [[x > 0]](FUN y -> IF y THEN [[2]](FUN w -> report w)
ELSE [[3]](FUN w -> report w))
= (FUN y -> IF y THEN [[2]](FUN w -> report w) ELSE
[[3]](FUN w -> report w)) (x > 0)
= (FUN y -> IF y THEN ((FUN w -> report w) 2) ELSE ((FUN
w -> report w) 3)) (x > 0)
• A function expression by itself does not get evaluated (well, it
gets turned into a closure), so it needs to be
handed to the continuation directly, except that, when it
eventually gets applied, it will need to additionally
take a continuation as another argument, and its body will need

to have been transformed with respect to this
additional argument. Therefore, we need to choose a new
continuation variable k to be the formal parameter
for passing a continuation into the function. Then, we need to
transform the body with k as its continuation,
and put it inside a continuation function with the same original
formal parameter together with k. The original
continuation κ is then applied to the result.
[[fn x => e]]κ = κ (FN x k -> [[e]]k) Where k is new (fresh for
κ)
Notice that we are not yet evaluating anything, so (FN x k ->
[[e]]k) is a CPS function expression, not
actually a closure.
Example:
[[fn x => x + 1]](FUN w -> report w)
= (FUN w -> report w) (FN x k -> (FUN y -> ((FUN z -> k (y +
z)) 1)) x)
• The CPS transformation for application mirrors its evaluation
order. In MicroML, we will uniformly use left-to-
right evaluation. Therefore, to evaluate an application, first
evaluate the function, e1, to a closure, then evaluate
e2 to a value to which that closure is applied. We create a new
continuation that takes the result of e1 and binds
it to v1, then evaluates e2 and binds it to v2. Finally, v1 is
applied to v2 and, since the CPS transformation makes
all functions take a continuation, it is also applied to the current
continuation κ. This rule is formalized by:
[[e1 e2]]κ = [[e1]](FUN v1 -> [[e2]](FUN v2 -> v1 v2 κ))
Where
v1 is fresh for e2 and κ
v2 is fresh for v1 and κ

Example: [[(fn x => x + 1) 2]](FUN w -> report w)
= [[(fn x => x + 1)]](FUN y -> [[2]](FUN z -> y z (FUN w ->
report w)))
= (FUN y -> [[2]](FUN z -> y z (FUN w -> report w)))
(FN x k -> (FUN a -> ((FUN b -> k (a + b)) 1)) x)
= (FUN y -> ((FUN z -> y z (FUN w -> report w)) 2))
(FN x k -> (FUN a -> ((FUN b -> k (a + b)) 1)) x)
5 Problem
1. (35 pts) Compute the following CPS transformation. All parts
should be transformed completely.
[[fn f => fn x => if x > 0 then f x else f ((-1) * x)]](FUN w ->
report w)
3

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