Ozyegin University CS 321 Programming Languages Sample Problems 05

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1 Page 1 of 7 Ozyegin University CS 321 Programming Languages Sample Problems (PLC Ex. 7.2.(i)) Write a C program containing a function void arrsum(int n, int arr[], int *sump) that computes and returns the sum of the first n elements of the given array arr. The result must be returned through the sump pointer. void arrsum(int n, int arr[], int *sump) { int sum = 0; for (int i = 0; i < n; i++) { sum += arr[i]; *sump = sum; 2. (PLC Ex. 7.2.(ii)) Write a C program containing a function void squares(int n, int arr[]) that, given n and an array arr of length n or more, fills arr[i] with i*i for i = 0,..., n 1. void squares(int n, int arr[]) { for (int i = 0; i < n; i++) { arr[i] = i * i; 3. Write a recursive C function void fib(int n, int *res) that computes the n th fibonacci number and returns it through the res pointer. void fib(int n, int *res) { if (n == 0 n == 1) { *res = 1; else { int f1; fib(n-1, &f1); int f2; fib(n-2, &f2); *res = f1 + f2; 4. Assuming that the environment and the store are initially empty, give a possible environment and store at the end of the following piece of C program.

2 CS 321 Sample Problems 05 Page 2 of 7 int n = 38; int a[3] = {5, 9, 13; int *p; p = &a[1]; *p = n; *(p+1) = a[0]*2; p = &n; p++; * ENV n: 70 a: 74 p: 75 * STORE In C++, parameters of functions that are declared using the & operator are passed by reference. What is the output of the C++-like program below? void main() { int a[3] = {3, 7, 10; int m = 56; int n = 99; mystery(m, n, a); print m, n, a[0], a[1], a[2] void mystery(int x, int &y, int a[]) { a[0] += 1; int temp = x; x = y; y = temp; An array can be represented as a pointer. For instance, the array definition int a[4] = {12, 13, 14, 15; can be represented using the following env/store: Env => a: 54

3 CS 321 Sample Problems 05 Page 3 of 7 Store => Explain how you can use this representation to find the length of an array. Find the difference between the address and the value of the array variable. That is, &a - a 7. A C-like program is given below. m = &n; // A *m = k[2]; // B k--; m++; // C *k = *m; // D m[2] = n; // E Starting from the env. and store given below, show the environment and the store after each statement. Store => After A: Store => After B: Store => After C: Store =>

4 CS 321 Sample Problems 05 Page 4 of 7 After D: Store => After E: Store => Which garbage collection algorithm would work the best for the following piece of Java program? Explain why. Remember that Strings in Java are immutable objects (i.e. once created, you may not modify them). String concat(string[] ss) { // Assume ss is very long String res = ""; for (int i = 0; i < ss.length; i++) res += ss[i]; return res; Suggest a better (i.e. more efficient) way to implement this code. Explain why it would be more efficient. This program creates a lot of short-lived objects. Each iteration of the loop creates a String object that becomes garbage in the next iteration. For this program, two space stop and copy probably works the best because this algorithm does not need to touch the many String objects that die young. Also, these young and dead objects may have caused fragmentation in the heap. Two space stop and copy fixes that problem as well. (Automatic) reference counting is likely to perform well, too. This is because there are no cyclic links between the dead objects. Furthermore, there are no links to other objects from the String objects; hence, deallocation of a String object does not cause a ripple effect. 9. Write a piece of program (in Java or C++) that would cause a memory leak if reference counting were used as the garbage collection technique. class Node { int item; Node next;

5 CS 321 Sample Problems 05 Page 5 of 7... Node node1 = new Node(); Node node2 = new Node(); node1.next = node2; node2.next = node1; node1 = null; node2 = null;... The code above creates a cycle between two objects, and then loses the pointers to these objects from the stack. 10. Implement a cycle function that takes a list lst and converts it to an infinite stream as if lst were a circular list. # let cycle lst = let rec helper items = match items with [] -> helper lst x::xs -> Cons(x, fun () -> helper xs) in helper lst val cycle : a list -> a stream = <fun> # let letters = cycle [ a ; b ; c ; d ];; val letters : char stream = Cons ( a, <fun>) # take 15 letters;; - : char list = [ a ; b ; c ; d ; a ; b ; c ; d ; a ; b ; c ; d ; a ; b ; c ] # take 10 (cycle [ a ]);; - : char list = [ a ; a ; a ; a ; a ; a ; a ; a ; a ; a ] 11. Implement the takewhile function for streams that takes a predicate p, a stream s, and returns as a list all the elements of s until there is an element of s that does not satisfy p. Extra: give an efficient tail-recursive solution. # let rec takewhile p s = if p (head s) then (head s) :: takewhile p (tail s) else [];; val takewhile : ( a -> bool) -> a stream -> a list = <fun> # takewhile (fun n -> n < 20) evens;; - : int list = [0; 2; 4; 6; 8; 10; 12; 14; 16; 18] # takewhile (fun n -> n < 30) primes;; - : int list = [2; 3; 5; 7; 11; 13; 17; 19; 23; 29] Note that the solution above is not tail-recursive.

6 CS 321 Sample Problems 05 Page 6 of Using the encodings below, show that mult 3 2 is 6. 0 = (λf.λx.x) 1 = (λf.λx.fx) 2 = (λf.λx.f(fx)) 3 = (λf.λx.f(f(fx))) mult = λm.λn.λf.λx.m(nf)x mult 3 2 = (λm.λn.λf.λx. m (nf) x) 3 2 (λn.λf.λx. 3 (nf) x) 2 λf.λx. 3 (2 f) x = λf.λx. 3 ((λf.λx.f(fx)) f) x expanded 2 = λf.λx. 3 ((λf.λx.f(fx)) f) x β-reducing underlined term λf.λx. 3 (λx.f(fx)) x = λf.λx. (λf.λx.f(f(fx))) (λx.f(fx)) x expanded 3 = λf.λx. (λf.λx.f(f(f x))) (λx.f(f x)) x β-reducing underlined term λf.λx. (λx.(λx.f(fx)) ((λx.f(fx)) ((λx.f(fx)) x))) x = λf.λx. (λx.(λx.f(f x)) ((λx.f(f x)) ((λx.f(f x)) x))) x β-reducing underlined term λf.λx. (λx.(λx.f(fx)) ((λx.f(fx)) (f(fx)))) x = λf.λx. (λx.(λx.f(f x)) ((λx.f(f x)) (f(f x)))) x β-reducing underlined term λf.λx. (λx.(λx.f(fx)) (f(f(f(fx))))) x = λf.λx. (λx.(λx.f(fx)) (f(f(f(fx))))) x β-reducing underlined term λf.λx. (λx.(f(f(f(f(f(fx))))))) x = λf.λx. (λx.(f(f(f(f(f(f x))))))) x β-reducing underlined term λf.λx.(f(f(f(f(f(fx)))))) = Reduce the following lambda expression to its normal form. (λf.λm.λp.m (f m p)) (λy.λz.y z) (λw.w)

7 CS 321 Sample Problems 05 Page 7 of 7 (λf.λm.λp.m (f m p)) (λy.λz.y z) (λw.w) (λm.λp.m ((λy.λz.y z) m p)) (λw.w) λp.(λw.w) ((λy.λz.y z) (λw.w) p) λp.((λy.λz.y z) (λw.w) p) λp.((λz.(λw.w) z) p) λp.((λw.w) p) λp.p