Homework Assignment 4 Solution

Description

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:

<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:

<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 “

• 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:

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):

Ok (ListType IntType)

 

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
	#	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

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