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Homework Assignment 4 Solution
• Assignment Policies
Collaboration Policy. It is acceptable for students to collaborate in understanding the material but not in solving the problems or programming. Use of the Internet is allowed, but should not include searching for existing solutions.
Under absolutely no circumstances code can be exchanged between students from di erent groups. Excerpts of code presented in class can be used.
• Assignment
In this assignment you are asked to extend the type-checker for the language REC with references, pairs, lists and trees. The interpreter is provided for you. This assignment is organized in four parts:
1. Part 1. Add type-checking for references.
2. Part 2. Add type-checking for pairs.
3. Part 3. Add type-checking for lists.
4. Part 4. Add type-checking for trees.
The new language including all these constructs will be called CHECKED PLUS.
• Type-Checking References
One new grammar production, which is added to the parser for CHECKED PLUS, is needed for the concrete syntax of expressions.
<Expression> ::= ()
The concrete syntax of types must be extended so as to allow the typing rules described below to be implemented. Two new productions are added, the one for unit and for ref:
1
<Type>
::=
int
<Type>
::=
bool
<Type>
::=
unit
<Type>
::= ref(<Type>)
<Type>
::=
(<Type> -> <Type>)
The typing rules are:
‘ e :: t
‘ newref(e) :: ref(t)
‘ e :: ref(t)
‘ deref(e) :: t
‘ e1 :: ref(t) ‘ e2 :: t
‘ setref(e1,e2) :: unit
‘ () :: unit
Note the use of the type unit, to indicate that the return result of the assignment operation is not important.
3.1 Task
You have to update the code for the type checker by adapting the le checker.ml of the
CHECKED language to EXPLICIT-REFS so that type_of_expr can handle :
let rec t y p e _ o f _ e x p r : expr -> texpr t e a _ r e s u l t = function
• ...
| Unit -> return UnitType
• ...
| NewRef ( e ) -> error " I m pl em e nt me!"
• | DeRef ( e ) -> error " I mp l em en t me!"
| SetRef ( e1 , e2 ) -> error " Im p le me n t me!"
Here are some examples:
1
#
chk
" let
x
=
newref (0) in deref (x)" ;;
-
: texpr Checked . ReM . result = Ok IntType
3
#
chk
" let
x
=
newref (0) in x" ;;
-
: texpr Checked . ReM . result = Ok ( RefType IntType )
5
#
chk
" let
x
=
newref (0) in setref (x ,4)" ;;
-
: texpr Checked . ReM . result = Ok UnitType
7
#
chk
" newref ( newref ( zero ?(0))) " ;;
-
: texpr Checked . ReM . result = Ok ( RefType ( RefType BoolType ))
9
#
chk
" let
x
=
0 in setref (x ,4) " ;;
-
: texpr
Checked . ReM . result = Error " setref : Expected a re f er en ce type "
• Pairs
4.1 Concrete syntax
Two new productions are added to the grammar of CHECKED PLUS:
2
<Expression> ::= pair(<Expression>,<Expression>)
<Expression> ::= unpair (<Identi er>,<Identi er>)=<Expression> in <Expression>
Some examples of programs using pairs follow:
pair (3 ,4)
2
pair ( pair (3 ,4) ,5)
4
pair ( zero ?(0) ,3)
6
pair ( proc ( x : int ) { x - 1 } , 4)
8
proc ( z : < int * int >) { unpair (x , y )= z in x }
10
proc ( z : < int * bool >) { unpair (x , y )= z in pair (y , x ) }
12
let f = proc ( z : < int * bool >) { unpair (x , y )= z in pair (y , x ) }
14 in ( f pair (1 , zero ?(0)))
Note that the concrete syntax of the types is also extended with a new production:
<Type>
::=
int
<Type>
::=
bool
<Type>
::=
unit
<Type>
::= ref(<Type>)
<Type>
::=
(<Type> -> <Type>)
<Type>
::=
<<Type> * <Type>>
Regarding the behavior of these expressions they are clear. For example, the expression (proc (z:<int*int>) unpair (x,y)=z in x pair(2, 3)) is a function that given a pair of integers, projects the rst component of the pair.
4.2 Task
Add two new cases to the de nition of the function type_of_expr in the le checker.ml:
let rec t y p e _ o f _ e x p r : expr -> texpr t e a _ r e s u l t = function
• ...
|
Pair ( e1 , e2 ) -> error " Im p le me n t
me!"
4
|
Unpair ( id1 , id2 , e1 , e2 ) -> error
" I mp le m en t me!"
You rst need to devise the appropriate typing rules!
Here are some examples:
#
chk " pair (2 , 3)" ;;
2
-
: texpr Checked . ReM . result = Ok ( PairType ( IntType , IntType ))
#
chk "( proc (z: < int *int >) { unpair (x,y)=z
in x} pair (2 ,
3)) ";;
4
-
: texpr Checked . ReM . result = Ok IntType
#
chk "( proc (z:int ) { unpair
(x,y)=z
in x}
pair (2 , 3)) " ;;
6
-
: texpr Checked . ReM . result
= Error
" unpair : Expected a
pair type "
3
• Lists
5.1 Concrete syntax
The concrete syntax for expressions should be extended with the following productions:
<Expression>
::=
emptylist <Type>
<Expression>
::= cons (<Expression>, <Expression>)
<Expression>
::= null? (<Expression>)
<Expression>
::= hd (<Expression>)
<Expression>
::=
tl (<Expression>)
The concrete syntax for types includes one new production (the last one listed below):
<Type>
::=
int
<Type>
::=
bool
<Type>
::=
unit
<Type>
::= ref(<Type>)
<Type>
::=
(<Type> -> <Type>)
<Type>
::=
<<Type> * <Type>>
<Type>
::=
list(<Type>)
The new type constructor is for typing lists. For example,
1. list(int) is the type of lists of integers
2. (int -> list(int)) is the type of functions that given an integer produce a list of integers.
Here are some sample expressions in the extended language. They are supplied in order to help you understand how each constructor works.
#
chk
" em pt y li st int " ;;
2
-
: texpr Checked . ReM . result = Ok ( ListType IntType )
4
#
chk
" cons (1 , em pt y li st
int )" ;;
-
: texpr Checked . ReM . result
= Ok ( ListType
IntType )
6
#
chk
"hd( cons (1 ,
e mp t yl is t
int ))" ;;
8
-
: texpr Checked . ReM . result = Ok IntType
10
#
chk
"tl( cons (1 ,
e mp t yl is t
int ))" ;;
-
: texpr Checked . ReM . result
= Ok ( ListType
IntType )
12
#
chk
" cons ( null ?( e mp ty l is t
int ) , e mp ty l is t int )" ;;
14
-
: texpr Checked . ReM . result =
Error
" cons : type
of
head
and tail do not
match "
16
#
chk
" proc (x: int )
{
proc
(l: list (int )) {
cons (x,l)
} }";;
18
-
: texpr Checked . ReM . result =
Ok
( FuncType ( IntType , FuncType ( ListType
IntType ,
ListType IntType )))
Here are the typing rules:
4
5.2 Typing rules
‘ e1 :: t
‘ e2 :: list(t)
‘ emptylist t :: list(t)
‘ cons(e1,e2) :: list(t)
‘ e :: list(t)
‘ e :: list(t)
‘ tl(e) :: list(t)
‘ hd(e) :: t
‘ e :: list(t)
‘ null?(e) :: bool
5.3 Task
Extend the type-checker to del with the new constructs:
• let rec t y p e _ o f _ e x p r : expr -> texpr t e a _ r e s u l t = function
...
3
|
E mp t yL is t ( t )
->
error
" I mp le m en t
me!"
|
Cons ( he , te )
->
error
" I m pl em e nt
me!"
• | Null ( e ) -> error " I mp le m en t me!"
• Hd ( e ) -> error " Im pl e me nt me!"
7| Tl ( e ) -> error " Im pl e me nt me!"
• Trees
6.1 Concrete syntax
The concrete syntax for expressions should be extended with the following productions:
<Expression>
::=
emptytree <Type>
<Expression>
::= node (<Expression>, <Expression>, <Expression>)
<Expression>
::= nullT? (<Expression>)
<Expression>
::= getData (<Expression>)
<Expression>
::= getLST (<Expression>)
<Expression>
::=
getRST (<Expression>)
The concrete syntax for types includes one new production (the last one listed below):
<Type> ::=
<Type> ::=
<Type> ::=
<Type> ::=
<Type> ::=
<Type> ::=
<Type> ::=
<Type> ::=
int
bool
unit
ref(<Type>)
(<Type> -> <Type>)
<<Type> * <Type>>
list(<Type>)
tree (<Type>)
Here is a sample program that assumes you have implemented the append operation on lists. Note that you will not be able to run this program unless you have extended the
5
interpreter to deal with lists and trees. This assignment only asks you to write the type-checker, not the interpreter. You are, of course, encouraged to write the interpreter too!
Here is another example. It should type-check with result and
evaluate to Ok (ListVal [NumVal 1; NumVal 2; NumVal 3]):
letrec append ( xs : list ( int )): list ( int ) -> list ( int ) =
• proc ( ys : list ( int )) {
if null ?( xs )
4
then
ys
else
cons ( hd ( xs ) ,(( append tl ( xs )) ys ))
• }
in
• letrec inorder ( x : tree ( int )): list ( int ) =
if nullT ?( x )
10 then e mp ty li s t int
else (( append ( inorder getLST ( x ))) cons ( getData ( x ) ,
12 ( inorder getRST ( x ))))
in
14 ( inorder node (2 ,
node (1 , e mp t yt re e
int , e m pt yt r ee
int ) ,
16
node (3 , e mp t yt re e
int , e m pt yt r ee
int )))
6.2 Typing rules
‘ emptytree t :: tree(t)
‘ e :: tree(t)
‘ getData(e) :: t
‘ e :: tree(t)
‘ getLST(e) :: tree(t)
‘ e1 :: t ‘ e2 :: tree(t) ‘ e3 :: tree(t)
‘ node(e1,e2,e3) :: tree(t) ‘ e :: tree(t)
‘ nullT?(e) :: bool ‘ e :: tree(t)
‘ getRST(e) :: tree(t)
6.3 Task
Extend the type-checker to del with the new constructs:
let rec t y p e _ o f _ e x p r : expr -> texpr t e a _ r e s u l t = function
• ...
|
E mp t yT re e ( t ) -> error " I mp le m en t me!"
4
|
Node ( de ,
le ,
re )
-> error " Im pl e me nt me!"
|
NullT ( t )
->
error
" I mp l em en t me!"
• | GetData ( t ) -> error " Im pl e me nt me!"
• GetLST ( t ) -> error " I m pl em e nt me!"
8| GetRST ( t ) -> error " I m pl em e nt me!"
Here are some sample expressions in the extended language. They are supplied in order you to help you understand how each construct works.
#
chk " em pt y tr ee int
" ;;
2
-
: texpr
Checked . ReM
. result =
Ok
( TreeType
IntType )
4
#
chk " nullT ?( node (1 ,
node (2 ,
e mp t yt re e int ,
e mp ty t re e int ) , e mp ty t re e int ))" ;;
-
: texpr
Checked . ReM
. result
=
Ok
BoolType
6
6
#
chk
" getData ( node (1 , node (2 ,
e m pt yt r ee int , e mp t yt re e int ) , e mp t yt re e
int ))"
;;
8
-
: texpr
Checked . ReM . result
= Ok IntType
10
#
chk
" getLST ( node (1 ,
node (2 ,
e mp t yt re e int , e mp ty t re e int ) , e mp ty t re e
int ))" ;;
-
: texpr
Checked . ReM
. result
=
Ok ( TreeType IntType )
• Submission instructions
Submit a le named HW3 <SURNAME>.zip through Canvas which includes all the source les required to run the interpreter and type-checker. Your grade will be determined as follows, for a total of 100 points:
Section
Grade
Unit
5
Reference
15
Pair
20
List
30
Tree
30
7
