上海交通大学试卷 ( A 卷 ) ( 2015 至 2016 学年第 2 学期 )

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1 上海交通大学试卷 ( A 卷 ) ( 2015 至 2016 学年第 2 学期 ) 班级号 (class) 学号 (id) 姓名 (name) 课程名称 程序语言理论 Theory of Programming Languages 成绩 (score) 1. (14 points) (a) Please give the value of a0, a1 after running the following Ocaml program: let f0 x = x + x + x let x0 = ref 3 let a0 = f0 (x0:=!x0+1;!x0) let f1 x = x() + x() + x() let x1 = ref 3 let a1 = f1 (fun () -> (x1:=!x1+1;!x1)) (b) Please argue the differences between call-by-value and call-by-name strategies. (c) Can you give one piece of Ocaml code that diverges(does not terminate) by using call-by-value strategy, but terminates by using call-by-name strategy? (d) Assume we have the following evaluation rules for untyped λ-calculus. Please select the rules for call-by-value strategy; and select those for call-by-name strategy. β-red1 (λx.t1 ) t 2 [x t 2 ] t 1 β-red2 (λx.t) v [x v] t App1 t 1 t 1 t 1 t 2 t 1 t 2 App2 t 2 t 2 t 1 t 2 t 1 t 2 App3 t 2 t 2 v 1 t 2 v 1 t 2 (a) a0 = 12; a1 = 15 (b) Call-by-name strategy always carries out the leftmost and outermost β-redex; Call-by-value strategy further requires that β-reduction happens after the argument part has been evaluated to a value; (c) let f x = 0 let rec g x = g x let a = f (g 0) (d) Call-by-name: β-red1, App1 Call-by-value: β-red2, App1, App3 1

2 我承诺, 我将严格遵守考试纪律 承诺人 : 题号得分批阅人 ( 流水阅卷教师签名处 ) 总分 2. (12 points) 2. (12 ) Given This problem the subtyping considers rules this in Ruby Fig. class: 1 on page 9. class A (a) How many attr_accessor different supertypes :x does {a : Top, b : Top} have? (b) Can you find m1 an infinite descending chain in the subtype relation? i.e., an infinite sequence of types = 4 0, S 1,... such that for any i, S i+1 <: S i. (c) What about end an infinite ascending chain? (d) If we add a new m2 typing rule m1 > 4 please figure end out the relation between Nat <: Float m3 = 4 1 = {a : Nat Float, b : {}} > 4 T end 2 = {a : Float Nat, b : {c : Nat}} If T 1 an T 2 has m4 subtyping relation, give the derivation tree; If not, explain the reason briefly. self.x = > 4 end (a) {a:top,b:top},{b:top,a:top},{a:top},{b:top},{}, Top end (b) S 0 ={}, S 1 ={a:top}, S 2 ={a:top,b:top}, etc. (c) T 0 = S 0 Top, T 1 = S 1 Top, T 2 = S 2 Top, etc. 1) (4 ) Is it possible to ine a class B such that evaluating B.new.m2 causes the method m2 (d) T 2 <: T ined 1. The derivation tree is omitted. in class A (not an override of m2) to return true? If so, ine class B as such, else explain why it is not possible. 2) (4 ) Is it possible to ine a class B such that evaluating B.new.m3 causes the method m3 ined in class A (not an override of m3) to return true? If so, ine class B as such, else explain why it is not possible. 3) (4 ) Is it possible to ine a class B such that evaluating B.new.m4 causes the method 2 m4 ined in class A (not an override of m4) to return true? If so, ine class B as such, else explain why it is not possible.

3 A Name:, Id: 3. (14 points) Recursive type is very powerful, it can make infinite data. This problem consider a typed λ-calculus with variant types, pairs, recursive types, fix, Unit, and Nat. Please refer to Fig. 2 on page 9 for syntax of variants. (a) Define a type NatList for a list of natural numbers where each list is either a nil of unit or cons of a pair of type Nat NatList. (b) Give a value of type NatList that has two elements 0 and succ0, and 0 is the head of this list. (c) Define a function hd of type NatList Nat that returns the head of a list. Return 0 if the given list is nil. (d) Define a function sum of type NatList Nat that returns the summation of all the numbers in a list. You can use (plus m n) directly for m + n. (a) NatList = µx. < nil : unit, cons : Nat X >. (b) < cons = (succ0, < cons = (0, < nil = unit > as NatList) > as NatList) > as NatList (c) hd = λl : NatList.case l of < nil = x > 0 < cons = p > p.1 (d) sum = fix λm : Natlist Nat.λl : NatList.case l of < nil = x > 0 < cons = p > plus p.1 (m p.2) 3

4 4. (10 points) Suppose a generating function F on the universe {a, b, c, d} is ined by the following inference rules: c d b c b d c a A set S is F -closed if F (S) S; it is F -consistent if S F (S); it is a fixed point if S = F (S). (a) List all the F -closed sets. (b) List all the F -consistent sets. (c) What is the least fixed point of F? What is the greatest fixed point of F? (a) F -closed: {b}, {a, b}, {a, b, c, d} (b) F -consistent sets:, {b}, {c, d}, {b, c, d}, {a, b, c, d} (c) µ F = {b}, ν F = {a, b, c, d}. 4

5 A Name:, Id: 5. (12 points) This problem uses System F extended with addition. Please refer to Fig. 3 on page 10 for initions of System F. (a) Give the appropriate System F typing rule(s) and evaluation rule(s) for addition expressions of the form t 1 + t 2. (This should be easy.) (b) Consider a typing context Γ = x : X.X X: i. What does T need to be for the program fragment x [T] (λy : Nat.y + 7) 11 to typecheck? ii. Given your choice for T, what type does x [T] (λy : Nat.y + 7) have? Give a typing derivation for Γ x [T] (λy : Nat.y + 7). (c) Given your choice for T. If v is an arbitrary value of type X.X X, then what might v [T] (λy : Nat.y + 7) 11 evaluate to? (d) If v is an arbitrary value of type (Nat Nat) Nat Nat then what might v (λy : Nat.y + 7) 11 evaluate to? (a) E-Add-L t 1 t 1 t 1 + t 2 t 1 + t 2 E-Add-R t 2 t 2 v 1 + t 2 v 1 + t 2 T-Add Γ t 1 : Nat, Γ t 2 : Nat Γ t 1 + t 2 : Nat (b) i. (Nat Nat) ii. Nat Nat. The derivation tree omitted. (c) It will always evaluate to 18 (or you can write 11+7) due to parametricity. In System F, any value of type X.X X is totally equivalent to the identity function. You can not have any computations on a value of X. (d) It could produce any value whatsoever. 5

6 6. (12 points) Assume we have a typed programming language formally ined by a small-step operational semantics and a typing judgment. Assume the Preservation and Progress Theorems hold for this language. Preservation: If t : T and t t, then t : T. Progress: If t : T for some T. Then either t is a value or else there is some t with t t. Consider each question below separately and explain your answers briefly. (a) Suppose we change the operational semantics by adding a new inference rule. i. Is it possible that Preservation does not hold any more? ii. Is it possible that Progress does not hold any more? (b) Suppose we change the type system by adding a new typing rule. i. Is it possible that Preservation does not hold any more? ii. Is it possible that Progress does not hold any more? (a) (b) i. Yes, the new rule might produce an ill-typed term. ii. No, more operational rules cannot make it harder to step. i. Yes, the new rule might let some term type-check that can take a step to produce a term that doesn t type-check. As an example, suppose we have a rule give (3+4)+() type Nat. It can step to 7+(), which does not type-check. ii. Yes, the new rules could allow a stuck term to type-check. For example, have a rule that gives 3 + () type Nat. 6

7 A Name:, Id: 7. (10 points) This problem considers to translate a λ-term to the corresponding de Bruijn term. Here are some initions for your reference. A naming context Γ = x n, x n 1,, x 1, x 0 assigns to each x i the de Bruijn index i. The d-place shift of a term t about cutoff c, written d c (t) is ined as d c (k) = { k if k < c k + d if k c d c (λ.t 1 ) = λ. d c+1 (t 1 ) d c (t 1 t 2 ) = d c (t 1 ) d c (t 2 ) The substitution of a term s for variable number j in a term t, written [j s]t, is ined as follows: { s if k = j [j s]k = k otherwise [j s](λ.t 1 ) = λ.[j s]t 1 [j s](t 1 t 2 ) = [j s]t 1 [j s]t 2 The β-reduction rule on de Bruijn terms is: (λ.t 1 ) t 2 1 [0 1 t 2 ]t 1 (a) Let Γ = wa, give the nameless representation of (λb.w (λa.b a)) (λb.b a). (b) Show the computation (β-reduction) of this de Bruijn term step by step. (a) Nameless representation: (λ.2 (λ.1 0)) (λ.0 1) (b) (λ.2 (λ.1 0)) (λ.0 1) 1 [0 1 (λ.0 1)](2 (λ.1 0)) = 1 [0 (λ.0 2)](2 (λ.1 0)) = 1 (2 [0 (λ.0 2)](λ.1 0))) = 1 (2 λ.[1 1 (λ.0 2)](1 0))) = 1 (2 λ.[1 (λ.0 3)](1 0))) = 1 (2 λ.((λ.0 3) 0)) = (1 λ.((λ.0 2) 0)) 7

8 8. (16 points) This problem consider a simple programming language on booleans. Syntax: t::= true false if t then t else t diverg v::= true false Evaluation: E-IfTrue if true then t2 else t 3 t 2 E-IfFalse if false then t2 else t 3 t 3 E-If t 1 t 1 if t 1 then t 2 else t 3 if t 1 then t 2 else t 3 E-Diverg diverg diverg Below is one possible encoding from this small language to untyped call-by-value λ- calculus. true 1 false 1 if t_1 then t_2 else t_3 1 diverg 1 = λx.λy.x (1) = λx.λy.y (2) = t 1 1 t 2 1 t 3 1 (3) = (λx.xx)(λx.xx) (4) An encoding is termination preserving if for any term t, t terminates if and only if t terminates. (a) Give a counter example to show that the above encoding 1 is NOT termination preserving; (b) Give a termination preserving encoding 2 from this small language to untyped call-by-value λ-calculus, and prove it. (c) If we want to keep the encoding 1, please modify the evaluation rules for this small language to make 1 termination preserving, and prove it. (a) The term t = if false then diverg else false terminates. However the encoding t 1 = (λx.λy.y)(ω)(λx.λy.y) will diverge under call-by-value strategy. (b) We can solve this problem by thunking E2 and E3. Here is one possible solution: Let I = λx.x ture 2 false 2 if t_1 then t_2 else t_3 2 diverg 2 = λx.λy.xi = λx.λy.yi = t 1 2 (λx. t 2 2 ) (λx. t 3 2 ) = (λx.xx)(λx.xx) 8

9 (c) We keep rules E-IfTrue, E-IfFalse and Diverg, and change rule E-If to the following three rules. t 1 t 1 E-If-1 if t 1 then t 2 else t 3 if t 1 then t 2 else t 3 E-If-2 t 2 t 2 if v 1 then t 2 else t 3 if v 1 then t 2 else t 3. t 3 t 3 E-If-1 if v 1 then v 2 else t 3 if v 1 then v 2 else t 3 Both proofs can be carried out by structural induction on terms. 9

10 A References 16.1 Algorithmic Subtyping 211 Figure 1: Subtype relation with records {} <: Extends 15-1 and 15-3 S <: S S<:U U<:T S<:T S <: Top (S-Refl) (S-Trans) (S-Top) T 1 <:S 1 S 2 <:T 2 S 1 S 2 <:T 1 T 2 {l i i 1..n } {k j j 1..m } k j =l i impliess j <:T i {k j :S j j 1..m }<:{l i :T i i 1..n } (S-Arrow) (S-Rcd) 136 Figure 16-1: Subtype relation with records (compact version) 11 Simple Extensions Figure 2: Variants <> Extendsλ (9-1) derivations for records involving combinations of depth, width, and permutation subtyping. Before we can drop S-Trans, we must first add a rule that New syntactic forms bundles depth, width, and permutation subtyping t 0 t into 0 one: t ::=... terms: (E-Case) i 1..n i 1..n j 1..m case t {l i } {k j } k j =l i impliess 0 of <l i =x <l=t> as T tagging j <:T i > t i i case t (S-Rcd) i 1..n j 1..m i 1..n 0 case tof <l i =x i > t i {k j :S case j }<:{l i :T i } of<l i 1..n i=x i > t i t i t i T ::= Lemma: If S <: T is derivable types: from the subtyping rules including S-RcdDepth, <l i =t i > as T <l i =t (E-Variant) i 1..n i > as T <l i :T i S-Rcd-Width, > type and of S-Rcd-Perm variants (but not S-Rcd), then it can also be derived using S-Rcd (and not S-RcdDepth, New typing S-Rcd-Width, rules or S-Rcd-Perm), Γand t:t vice New evaluation rules versa. t t Γ t j :T j i 1..n i 1..n i 1..n case (<l j =v j > ast) Proof: of <l Straightforward i =x i > t i induction Γ <l on derivations. j =t j >as <l i :T i >:<l i :T i > [x j v j ]t j (T-Variant) Lemma justifies eliminating rules S-RcdDepth, S-Rcd-Width, and (E-CaseVariant) i 1..n S-Rcd-Perm in favor of S-Rcd. Figure 16-1 Γ t summarizes 0 :<l i :T i the> resulting system. Next, we show that, in the systemfor of eachi Figure 16-1, Γ,x i the :T i reflexivity t i :T and transitivity rules are inessential. Γ case t 0 of <l i =x i > t i (T-Case) i 1..n :T Lemma: Figure 11-11: Variants 1. S<:S can be derived for every type S without using S-Refl. 2. If S<:T is derivable, then it can be derived without using S-Trans Variants Proof: Exercise [Recommended, ]. Binary sums generalize to labeled variants just as products generalize to labeled records. [ ]: IfInstead we addoft the 1 type +T 2, we Bool, write<l how do 1 :Tthese 1,l 2 :T properties 2 >, wherel change? 1 andl 2 are Exercise field labels. Instead of inl tas T 1 +T 2, we write <l 1 =t>as <l 1 :T 1,l 2 :T 2 >. This Andbrings insteadusof tolabeling the inition the branches of the algorithmic of the casesubtype with inl relation. and inr, we use the same labels as the corresponding sum type. With these generalizations, Definition: The algorithmic subtyping relation is the least relation on types thegetaddr example from the previous section becomes: closed under the rules in Figure Addr = <physical:physicaladdr, virtual:virtualaddr>; a = <physical=pa> as Addr; a : Addr 10 getname = λa:addr. case a of <physical=x> x.firstlast <virtual=y> y.name;

11 23.3 System F 343 Figure 3: System F Based onλ (9-1) Syntax t ::= x λx:t.t t t λx.t t [T] terms: variable abstraction application type abstraction type application Evaluation t t t 1 t 1 t 1 t 2 t 1 t 2 t 2 t 2 v 1 t 2 v 1 t 2 (E-App1) (E-App2) (λx:t 11.t 12 ) v 2 [x v 2 ]t 12 (E-AppAbs) v ::= λx:t.t λx.t values: abstraction value type abstraction value t 1 t 1 t 1 [T 2 ] t 1 [T 2] (E-TApp) T ::= X T T X.T types: type variable type of functions universal type (λx.t 12 )[T 2 ] [X T 2 ]t 12 (E-TappTabs) Typing x:t Γ Γ x:t Γ t:t (T-Var) Γ ::= Γ,x:T Γ,X contexts: empty context term variable binding type variable binding Γ,x:T 1 t 2 :T 2 Γ λx:t 1.t 2 :T 1 T 2 Γ t 1 :T 11 T 12 Γ t 2 :T 11 Γ t 1 t 2 :T 12 (T-Abs) (T-App) Γ,X t 2 :T 2 Γ λx.t 2 : X.T 2 (T-TAbs) Γ t 1 : X.T 12 Γ t 1 [T 2 ]:[X T 2 ]T 12 (T-TApp) Figure 23-1: Polymorphic lambda-calculus (System F) is to keep track of scopes and make sure that the same type variable is not added twice to a context. In later chapters, we will annotate type variables with information of various kinds, such as bounds (Chapter 26) and kinds (Chapter 29). Figure 23-1 shows the complete inition of the polymorphic lambdacalculus, with differences fromλ highlighted. As usual, this summary de- 11 fines just the pure calculus, omitting other type constructors such as records,

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