# 9.8. Exercises¶

Solutions to exercises are available to students in Cornell’s CS 3110. Instructors at other institutions are welcome to contact Michael Clarkson for access.

Many of these exercises rely on the SimPL interpreter as starter code. You can download it here: simpl.zip.

Exercise: parse [★]

Run make utop in the SimPL interpreter implementation. It will compile the interpreter and launch utop. Evaluate the following expressions. Note what each returns.

• parse "22"

• parse "1 + 2 + 3"

• parse "let x = 2 in 20 + x"

Also evaluate these expressions, which will raise exceptions. Explain why each one is an error, and whether the error occurs during parsing or lexing.

• parse "3.14"

• parse "3+"

Exercise: simpl ids [★★]

Examine the definition of the id regular expression in the SimPL lexer. Identify at least one way in which it differs from the definition of OCaml identifiers.

Exercise: times parsing [★★]

In the SimPL parser, the TIMES token is declared as having higher precedence than PLUS, and as being left associative. Let’s experiment with other choices.

• Evaluate parse "1*2*3". Note the AST. Now change the declaration of the associativity of TIMES in parser.mly to be %right instead of %left. Recompile and reevaluate parse "1*2*3". How did the AST change? Before moving on, restore the declaration to be %left.

• Evaluate parse "1+2*3". Note the AST. Now swap the declaration %left TIMES in parser.mly with the declaration %left PLUS. Recompile and reevaluate parse "1+2*3". How did the AST change? Before moving on, restore the original declaration order.

Exercise: infer [★★]

Type inference for SimPL can be done in a much simpler way than for the larger language (with anonymous functions and let expression) that we considered in the section on type inference.

Run make in the SimPL interpreter implementation. It will compile the interpreter and launch utop. Now, define a function infer : string -> typ such that infer s parses s into an expression and infers the type of s in the empty context. Your solution will make use of the typeof function. You don’t need constraint collection or unification.

Try out your infer function on these test cases:

• "3110"

• "1 <= 2"

• "let x = 2 in 20 + x"

Exercise: subexpression types [★★]

Suppose that a SimPL expression is well typed in a context ctx. Are all of its subexpressions also well typed in ctx? For every subexpression, does there exist some context in which the subexpression is well typed? Why or why not?

Exercise: typing [★★]

Use the SimPL type system to show that {} |- let x = 0 in if x <= 1 then 22 else 42 : int.

Exercise: substitution [★★]

What is the result of the following substitutions?

• (x + 1){2/x}

• (x + y){2/x}{3/y}

• (x + y){1/z}

• (let x = 1 in x + 1){2/x}

• (x + (let x=1 in x+1)){2/x}

• ((let x=1 in x+1) + x){2/x}

• (let x=y in x+1){2/y}

• (let x=x in x+1){2/x}

Exercise: step expressions [★]

Here is an example of evaluating an expression:

  7+5*2
-->  (step * operation)
7+10
-->  (step + operation)
17


There are two steps in that example, and we’ve annotated each step with a parenthetical comment to hint at which evaluation rule we’ve used. We stopped evaluating when we reached a value.

Evaluate the following expressions using the small-step substitution model. Use the “long form” of evaluation that we demonstrated above, in which you provide a hint as to which rule is applied at each step.

• (3 + 5) * 2 (2 steps)

• if 2 + 3 <= 4 then 1 + 1 else 2 + 2 (4 steps)

Exercise: step let expressions [★★]

Evaluate these expressions, again using the “long form” from the previous exercise.

• let x = 2 + 2 in x + x (3 steps)

• let x = 5 in ((let x = 6 in x) + x) (3 steps)

• let x = 1 in (let x = x + x in x + x) (4 steps)

Exercise: variants [★]

Evaluate these Core OCaml expressions using the small-step substitution model:

• Left (1+2) (1 step)

• match Left 42 with Left x -> x+1 | Right y -> y-1 (2 steps)

Exercise: application [★★]

Evaluate these Core OCaml expressions using the small-step substitution model:

• (fun x -> 3 + x) 2 (2 steps)

• let f = (fun x -> x + x) in (f 3) + (f 3) (6 steps)

• let f = fun x -> x + x in let x = 1 in let g = fun y -> x + f y in g 3 (7 steps)

• let f = (fun x -> fun y -> x + y) in let g = f 3 in (g 1) + (f 2 3) (9 steps)

Exercise: omega [★★★]

Try evaluating (fun x -> x x) (fun x -> x x). This expression, which is usually called $$\Omega$$, doesn’t type check in real OCaml, but we can still use the Core OCaml small-step semantics on it.

Exercise: pair parsing [★★★]

Add pairs (i.e., tuples with exactly two components) to SimPL. Implement lexing and parsing of pairs. Assume that the parentheses around the pair are required (not optional, as they sometimes are in OCaml). Follow this strategy:

• Add a constructor for pairs to the expr type.

• Add a comma token to the parser.

• Implement lexing the comma token.

• Implement parsing of pairs.

When you compile, you will get some inexhaustive pattern match warnings, because you have not yet implemented type checking nor interpretation of pairs. But you can still try parsing them in utop with the parse function.

Exercise: pair type checking [★★★]

Implement type checking of pairs. Follow this strategy:

• Write down a new typing rule before implementing any code.

• Add a new constructor for pairs to the typ type.

• Add a new branch to typeof.

Exercise: pair evaluation [★★★]

Implement evaluation of pairs. Follow this strategy:

• Implement is_value for pairs. A pair of values (e.g., (0,1)) is itself a value, so the function will need to become recursive.

• Implement subst for pairs: (e1, e2){v/x} = (e1{v/x}, e2{v/x}).

• Implement small-step and big-step evaluation of pairs, using these rules:

(e1, e2) --> (e1', e2)
if e1 --> e1'

(v1, e2) --> (v1, e2')
if e2 --> e2'

(e1, e2) ==> (v1, v2)
if e1 ==> v1
and e2 ==> v2


Exercise: desugar list [★]

Suppose we treat list expressions like syntactic sugar in the following way:

• [] is syntactic sugar for Left 0.

• e1 :: e2 is syntactic sugar for Right (e1, e2).

What is the core OCaml expression to which [1; 2; 3] desugars?

Exercise: list not empty [★★]

Write a core OCaml function not_empty that returns 1 if a list is non-empty and 0 if the list is empty. Use the substitution model to check that your function behaves properly on these test cases:

• not_empty []

• not_empty [1]

Exercise: list not empty [★★★★]

In core OCaml, there are only two patterns: Left x and Right x, where x is a variable name. But in full OCaml, patterns are far more general. Let’s see how far we can generalize patterns in core OCaml.

Step 1: Here is a BNF grammar for patterns, and slightly revised BNF grammar for expressions:

p ::= i | (p1, p2) | Left p | Right p | x | _

e ::= ...
| match e with | p1 -> e1 | p2 -> e2 | ... | pn -> en


In the revised syntax for match, only the very first | on the line, immediately before the keyword match, is meta-syntax. The remaining four | on the line are syntax. Note that we require | before the first pattern.

Step 2: A value v matches a pattern p if by substituting any variables or wildcards in p with values, we can obtain exactly v. For example:

• 2 matches x because x{2/x} is 2.

• Right(0,Left 0) matches Right(x,_) because Right(x,_){0/x}{Left 0/_} is Right(0,Left 0).

Let’s define a new ternary relation called matches, guided by those examples:

v =~ p // s


Pronounce this relation as “v matches p producing substitutions s.”

Here, s is a sequence of substitutions, such as {0/x}{Left 3/y}{(1,2)/z}. There is just a single rule for this relation:

v =~ p // s
if v = p s


For example,

2 =~ x // {2/x}
because 2 = x{2/x}


Step 3: To evaluate a match expression:

• Evaluate the expression being matched to a value.

• If that expression matches the first pattern, evaluate the expression corresponding to that pattern.

• Otherwise, match against the second pattern, the third, etc.

• If none of the patterns matches, evaluation is stuck: it cannot take any more steps.

Using those insights, complete the following evaluation rules by filling in the places marked with ???:

(* This rule should implement evaluation of e. *)
match e with | p1 -> e1 | p2 -> e2 | ... | pn -> en
--> ???
if ???

(* This rule implements moving past p1 to the next pattern. *)
match v with | p1 -> e1 | p2 -> e2 | ... | pn -> en
--> match v with | p2 -> e2 | ... | pn -> en
if there does not exist an s such that ???

(* This rule implements matching v with p1 then proceeding to evaluate e1. *)
match v with | p1 -> e1 | p2 -> e2 | ... | pn -> en
--> ??? (* something involving e1 *)
if ???


Note that we don’t need to write the following rule explicitly:

match v with |  -/->


Evaluation will get stuck at that point because none of the three other rules above will apply.

Step 4: Double check your rules by evaluating the following expression:

match (1 + 2, 3) with | (1,0) -> 4 | (1,x) -> x | (x,y) -> x + y

Exercise: let rec [★★★★]

One of the evaluation rules for let is

let x = v in e --> e{v/x}


We could try adapting that to let rec:

let rec x = v in e --> e{v/x}   (* broken *)


But that rule doesn’t work properly, as we see in the following example:

  let rec fact = fun x ->
if x <= 1 then 1 else x * (fact (x - 1)) in
fact 3

-->

(fun x -> if x <= 1 then 1 else x * (fact (x - 1)) 3

-->

if 3 <= 1 then 1 else 3 * (fact (3 - 1))

-->

3 * (fact (3 - 1))

-->

3 * (fact 2)

-/->


We’re now stuck, because we need to evaluate fact, but it doesn’t step. In essence, the semantic rule we used “forgot” the function value that should have been associated with fact.

A good way to fix this problem is to introduce a new language construct for recursion called simply rec. (Note that OCaml does not have any construct that corresponds directly to rec.) Formally, we extend the syntax for expressions as follows:

e ::= ...
| rec f -> e


and add the following evaluation rule:

rec f -> e  -->  e{(rec f -> e)/f}


The intuitive reading of this rule is that when evaluating rec f -> e, we “unfold” f in the body of e. For example, here is an infinite loop coded with rec:

  rec f -> f

-->  (* step rec *)

f{(rec f -> f)/f}

= (* substitute *)

rec f -> f

--> (* step rec *)

f{(rec f -> f)/f}

...


Now we can use rec to implement let rec. Anywhere let rec appears in a program:

let rec f = e1 in e2


we desugar (i.e., rewrite) it to

let f = rec f -> e1 in e2


Note that the second occurrence of f (inside the rec) shadows the first one. Going back to the fact example, its desugared version is

let fact = rec fact -> fun x ->
if x <= 1 then 1 else x * (fact (x - 1)) in
fact 3


Evaluate the following expression (17 steps, we think, though it does get pretty tedious). You may want to simplify your life by writing “F” in place of (rec fact -> fun x -> if x <= 1 then 1 else x * (fact (x-1)))

let rec fact = fun x ->
if x <= 1 then 1 else x * (fact (x - 1)) in
fact 3


Exercise: simple expressions [★]

In the small-step substitution model, evaluation of an expression was rather list-like: we could write an evaluation in a linear form like e --> e1 --> e2 --> ... --> en --> v. In the big-step environment model, evaluation is instead rather tree-like: evaluations have a nested, recursive structure. Here’s an example:

<{}, (3 + 5) * 2> ==> 16          (op rule)
because <{}, (3 + 5)> ==> 8   (op rule)
because <{},3> ==> 3      (int const rule)
and     <{},5> ==> 5      (int const rule)
and 3+5 is 8
and <{}, 2> ==> 2             (int const rule)
and 8*2 is 16


We’ve used indentation here to show the shape of the tree, and we’ve labeled each usage of one of the semantic rules.

Evaluate the following expressions using the big-step environment model. Use the notation for evaluation that we demonstrated above, in which you provide a hint as to which rule is applied at each node in the tree.

• 110 + 3*1000 hint: three uses of the constant rule, two uses of the op rule

• if 2 + 3 < 4 then 1 + 1 else 2 + 2 hint: five uses of constant, three uses of op, one use of if(else)

Exercise: let and match expressions [★★]

Evaluate these expressions, continuing to use the tree notation, and continuing to label each usage of a rule.

• let x=0 in 1 hint: one use of let, two uses of constant

• let x=2 in x+1 hint: one use of let, two uses of constant, one use of op, one use of variable

• match Left 2 with Left x -> x+1 | Right x -> x-1 hint: one use of match(left), two uses of constant, one use of op, one use of variable

Exercise: closures [★★]

Evaluate these expressions:

• (fun x -> x+1) 2 hint: one use of application, one use of anonymous function, two uses of constant, one use of op, one use of variable

• let f = fun x -> x+1 in f 2 hint: one use of let, one use of anonymous function, one use of application, two uses of variable, one use of op, two uses of constant

Exercise: lexical scope and shadowing [★★]

Evaluate these expressions:

• let x=0 in x + (let x=1 in x) hint: two uses of let, two uses of variable, one use of op, two uses of constant

• let x=1 in let f=fun y -> x in let x=2 in f 0 hint: three uses of let, one use of anonymous function, one use of application, two uses of variable, three uses of constant

Exercise: more evaluation [★★]

Evaluate these:

• let x = 2 + 2 in x + x

• let x = 1 in let x = x + x in x + x

• let f = fun x -> fun y -> x + y in let g = f 3 in g 2

• let f = fst (let x = 3 in fun y -> x, 2) in f 0

Exercise: dynamic scope [★★★]

Use dynamic scope to evaluate the following expression. You do not need to write down all of the evaluation steps unless you find it helpful. Compare your answer to the answer you would expect from a language with lexical scope.

let x = 5 in
let f y = x + y in
let x = 4 in
f 3


Exercise: more dynamic scope [★★★]

Use dynamic scope to evaluate the following expressions. Compare your answers to the answers you would expect from a language with lexical scope.

Expression 1:

let x = 5 in
let f y = x + y in
let g x = f x in
let x = 4 in
g 3


Expression 2:

let f y = x + y in
let x = 3 in
let y = 4 in
f 2


Exercise: constraints [★★]

Show the derivation of the env |- e : t -| C relation for these expressions:

1. fun x -> ( + ) 1 x
2. fun b -> if b then false else true
3. fun x -> fun y -> if x <= y then y else x


Exercise: unify [★★]

Use the unification algorithm to solve the following system of constraints. Your answer should be a substitution, in the sense that the unification algorithm defines that term.

X = int
Y = X -> X


Exercise: unify more [★★★]

Use the unification algorithm to solve the following system of constraints. Your answer should be a substitution, in the sense that the unification algorithm defines that term.

X -> Y = Y -> Z
Z = U -> W


Exercise: infer apply [★★★]

Using the HM type inference algorithm, infer the type of the following definition:

let apply f x = f x


Remember to go through these steps:

• desugar the definition entirely (i.e., construct an AST)

• collect constraints

• solve the constraints with unification

Exercise: infer double [★★★]

Using the HM type inference algorithm, infer the type of the following definition:

let double f x = f (f x)


Exercise: infer S [★★★★]

Using the HM type inference algorithm, infer the type of the following definition:

let s x y z = (x z) (y z)