20-CS Programming Languages Fall Final Exam! Answer all questions Be sure to put your name on the paper in the space provided!

Transcription

1 20-CS Programming Languages Fall 2018 Final Exam! Answer all questions Be sure to put your name on the paper in the space provided! Name Signature The code below compiles and runs just fine (worse that good, right?) class A { public String tostring (Object obj) { return obj.tostring(); class B extends A { public String tostring (Object obj) { return ((Integer)obj).toString(); public class prob1 { public static void main (String args[]) { LinkedList ll = new LinkedList(); ll.add(new Integer(10)); ll.add("flibbertigibbet"); System.out.println((new A()).toString(ll.get(0))); System.out.println((new B()).toString(ll.get(0))); System.out.println((new B()).toString(ll.get(1))); 1. Add a line to the main method that causes a runtime failure with a ClassCastException. Write the line below and connect it to a place in main by drawing a curvy line beginning at the start of what you wrote and ending in an arrow head between two lines in main. System.out.println((new B()).toString(ll.get(1))); 2. This code is an ultra simple, stripped down version of something that is common in practice: a developer has created class A and put it in a library and a second developer wants to extend class A with class B which, of course, is trouble. What can the first developer do to prevent the second developer from doing this successfully? Full details are required. class A { public final String tostring (Object obj) { return obj.tostring();

2 Streams can be implemented using semaphores and threads. See the last page for a reminder of what this entails for the classic producer-consumer problem. Streams can also be implemented like this: class Stream { int first; boolean empty = true; Stream rest () { return null; with sample producer-consumer application like this: class Producer extends Stream { Producer (int f) { first = f; empty = false; Stream rest () { return new Producer(first+1); class Consumer { public void consume (long n) { Stream p = new Producer(1); for (int i=0 ; i < n ; i++) { System.out.print(p.first+" "); p = p.rest(); System.out.println(); public class prob2 { public static void main (String args[]) { (new Consumer()).consume(100); 3. Which of the two Stream implementations will be faster and why? The Stream solution above is much faster because the overhead of threads is gone. 4. Name an application where the Threaded version of Stream will get into serious trouble and say what that trouble is. Hamming s sequence problem is in trouble because new threads are generated as the sequence gets larger. Eventually the threads exhaust computer resources.

3 The following code includes a procedure arraytocollection which is intended to copy items of an array of arbitrary type into a Collection of arbitrary type. In this particular instance the array is of type String and the Collection is of type Vector <String>. import java.util.*; class A { public Collection <?> arraytocollection (Object [] a, Collection <?> b) { for (int i=0 ; i < a.length ; i++) b.add(a[i]); return b; public void doit () { String arr[] = new String[10]; for (int i=0 ; i < 10 ; i++) arr[i] = new String(String.valueOf(i+3)); Vector <String> v = new Vector <String> (); Collection <?> w = arraytocollection(arr, v); System.out.println(w.toString()); public class prob3a { public static void main (String args[]) { (new A()).doit(); 5. What about arraytocollection causes this to fail to compile? Consider b.add[a(i)]. Object a[i] is of type Object but the Collection needs to be of type String. It is not allowed to convert Object to String. 6. Fix arraytocollection so that the above code compiles and runs successfully. That is, write the fixed first line of arraytocollection (public followed by return value, followed by arraytocollection followed by argument list) below. import java.util.*; class A { public <T> Collection <T> arraytocollection (T [] a, Collection <T> b) { for (int i=0 ; i < a.length ; i++) b.add(a[i]); return b; public void doit () { String arr[] = new String[10]; for (int i=0 ; i < 10 ; i++) arr[i] = new String(String.valueOf(i+3)); Vector <String> v = new Vector <String> (); Collection <?> w = arraytocollection(arr, v); System.out.println(w.toString());

4 The following procedure seq is intended to produce an infinite list of increasing numbers. (define seq (lambda (n) (cons n (seq (+ n 1))))) 7. Why is this useless? The procedure does not terminate when it is called 8. The following procedure qes is a fix. (define qes (lambda (n) (cons n (lambda () (qes (+ n 1)))))) Fill in the missing lines below to complete a procedure called take that outputs the first n numbers in (qes 1). (define take (lambda (n) (letrec ((t (lambda (m s) (if (zero? m) () (cons (car s) (t (- m 1) ((cdr s)))))))) (t n (qes 1)))))

5 The following procedure is an example of currying: a feature of functional languages like Haskell and Scheme: (define f (lambda (op) (lambda (a) (lambda (b) (op a b))))) Thus, (f +) is a procedure which adds 2 arguments, and ((f +) 4) is a procedure which adds 4 to a single argument. We can use the curry feature to write anonymous functions, even recursive ones. A first attempt at this for implementing factorial might be the following: (define g (lambda (op) (lambda (n) (if (zero? n) 1 (* n (op (- n 1))))))) The idea is that op represents the factorial function and, since g is being designed to build the anonymous factorial function, we should be able to pass g to itself and use it like this: ((g g) 10) to get factorial Why won t this work? g is not a factorial function but rather a maker of a factorial function. 10. What is the simple fix that will make this work? Change to (* n (op (- n 1))) (* n ((op op) (- n 1))) Then to get, say, factorial 10 do this: ((g g) 10).

6 11. What is the difference between a and b in the following: (define a (car (hello world I am safe))) (define b ((lambda (x) (car x)) (hello world I am safe))) There is no difference 12. How can this difference be used in an alternative to your solution in question 10 that results in a factorial specification that looks like a factorial specification? It is better to just write your solution below (which is almost the same as g fixed as requested) than to explain in English but if you need to use English, go ahead. (define g (lambda (op) ((lambda (fact) (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))) (lambda (x) ((op op) x))))) ;; this line become fact

7 13. The following two expressions have the same value which is [8,10,12,16]: map (*2) [4, 5, 6, 8] fmap (*2) [4, 5, 6, 8] So, what s the difference? A functor must implement fmap. It is not necessary to implement map (and I am not sure you could do so if you wanted to). 14. In Haskell underlying types such as Integer and [Char] are encased in types that are designed to have code operate safely in case of failure. The value of fmap (*2) [4] is [8]. Where or what is the encasing type and are there other such encasing types that you can remember (if so, name them)? This encasing type is []. Others would be Either, Maybe. In class we also considered State and Tree. 15. What is the type signature of fmap (*2)? You can do this but it might be a struggle. Use symbol f to represent the encasing type and f b to represent the encased underlying type. Use symbol -> to separate input from output, and use symbol => to separate typeclass information from argument and output types. All you need to do is figure out what type classes you need to specify (the left side of =>) and the correct arrangement of input and output types (the right side of =>). fmap (*2) :: (Num b, Functor f) => f b -> f b 16. Write a list comprehension that is equivalent to [1,2] >>= (\n -> [ a, b ] >>= (\ch -> return (n,ch))) I am asking because I am sure you understand list comprehensions so if you get this right it means you also understand the bind operator. [(n,ch) n <- [1,2], ch <- [ a, b ]]

8 17. Write an alternative but equivalent expression for return 4 >>= (\x -> (Just (*x))) >>= (\f -> (Just (f 3))) using applicative operators <$> and <*>. Again, I am just trying to determine whether you understand something about the bind operator. In case you have forgotten about the applicative operators: first, Monads must be applicative by definition, second these operators work almost like their underlying type counterparts. (*) <$> (Just 3) <*> (Just 4) 18. Why is return needed in the above? Wouldn t 4 >>=... suffice? Actually, return is not needed. The 4 needs to be lifted to encasement by a Maybe Monad in order for the bind operator (>>=) to be used correctly and this is done by return. But it could also be done as follows: Just 4 >>= (\x -> (Just (*x))) >>= (\f -> (Just (f 3))) 19. What is wrong with the following code? #include <stdio.h> #include <stdlib.h> int getauthorization (int p) { if (p < 1000) return 0; else return 1; int main (int argc, char **argv) { int n; unsigned short b; n = atoi(argv[1]); if (!getauthorization(n)) { printf("authorization disallowed\n"); exit(0); b = n; printf("authorization for slot %d allowed\n", b); In line b = n; conversion from int to unsigned short will produce a number b that is different from n. Crafting a particular n will result in a b that would have passed the getauthorization test. If n is large enough, it will pass the test but b will be the (illegal) slot granted to the user. 20. Provide a simple fix. int main (int argc, char **argv) { unsigned short b; b = atoi(argv[1]); if (!getauthorization(b)) { printf("authorization disallowed\n"); exit(0); printf("authorization for slot %d allowed\n", b);

9 import java.util.concurrent.semaphore; class Monitor { private int contents; private Semaphore tokenout = new Semaphore(0,true); private Semaphore tokenin = new Semaphore(1,true); public int get() throws InterruptedException { tokenout.acquire(); tokenin.release(); return contents; public void put(int value) throws InterruptedException { tokenin.acquire(); contents = value; tokenout.release(); class Producer extends Thread { private Monitor monitor; private int number; public Producer (Monitor c) { monitor = c; public void run() { for (int i = 1; ; i++) { try { monitor.put(i); catch (InterruptedException e) { class Consumer extends Thread { private Monitor monitor; private int number; private long iters; public Consumer (Monitor c, long n) { monitor = c; iters = n; public void run() { int value = 0; for (int i = 0 ; i < iters ; i++) { try { value = monitor.get(); catch (InterruptedException e) { System.out.print(value+" "); System.out.println(); System.exit(1); public class ref1 { public static void main (String args[]) { Monitor m = new Monitor(); Producer p = new Producer(m); Consumer c = new Consumer(m, 100); p.start(); c.start();