CS107 Handout 43 Spring 2007 June 8 th, 2007 CS107 Final Exam

Transcription

1 CS107 Handout 43 Spring 2007 June 8 th, 2007 CS107 Final Exam This is an open-note exam. You can refer to any course handouts, handwritten lecture notes, and printouts of any code relevant to a CS107 assignment. You may not use any laptops, cell phones, or handheld devices of any sort. Those taking the exam remotely should phone in if they have questions. Once you re done, fax the exam to Stanford, but hold on to the original until you get the graded fax copy back. Cell phone is , and fax number is leland username: Last Name: First Name: I accept the letter and spirit of the honor code. I ve neither given nor received aid on this exam. I pledge to write more neatly than I ever have in my entire life. (signed) Score Grader 1. Munchies (10) 2. All Things Scheme (15) 3. C versus Scheme (6) 4. Java Runtime (7) 5. Java Short Answer (7) 6. Conserving Threads (10) Total (55) SCPD students who want their exams sent back through regular mail, check here:

2 2 Problem 1: Munchies (10 points) Given the following C++ class definition, generate code for the munchy::cheeto method. Assume that the parameters have already been set up for you. Be clear about what code pertains to which line. Recall that C++ references are automatically dereferenced pointers, and k-argument methods are really (k + 1)-argument functions, where the address of the receiving object is passed in as the bottommost parameter. The address of the first instruction of the cheeto method is synonymous with <munchy::cheeto>. You have this and the next page for your code. class munchy { public: char& cheeto(munchy& pretzel, char *sunchip); ; private: short *pringle; short bugle[2]; char dorito[12]; line 1 line 2 char& munchy::cheeto(munchy& pretzel, char *sunchip) { *(munchy **)sunchip += sunchip[8]; return dorito[pretzel.cheeto(*this, sunchip + 1)];

3 (more space for Problem 1) 3

4 4 Problem 2: Scheme (15 points) a) (4 points) A derangement is a permutation of the integers 1 through n such that no number k falls in the k th position. So, ( ) is certainly a permutation, but it s not a derangement, because the 3 sits in the 3 rd position. ( ) is a derangement of length 6, and ( ) is a derangement of length 7. ( ) is as far as a permutation can be from a derangement. Write a function called derangement? that returns #t if and only if the supplied list of numbers, assumed to be a real permutation of all the integers 1 through the list length, is also a derangement. Use car-cdr recursion. Feel free to write helper functions. ;;; ;; Returns true if and only if the specified permutation [which ;; is assumed to be an actually permutation of all numbers between ;; 1 and (length permutation), inclusive] is also a derangement. ;; ;; You needn't confirm that the permutation is really a permutation ;; of the first (length permutation) numbers. Just assume it is. ;; (define (derangement? permutation) (

5 5 b) (3 points) Write a routine called mirror, which takes an arbitrary item and returns the recursively reversed version of the original item. There s a built-in function called reverse that takes a list and returns a new list where all the top-level items come in reverse, but there s no recursive reversing like we want from our mirror function. Your answer should be very short and must use mapping instead of exposed car-cdr recursion. ;;; ;; Returns a deep clone of the specified item, where ;; all lists, sublists, subsublists, and so forth, are reversed. ;; There is a built-in called reverse that reverses the order of the ;; top-level elements of a list. ;; ;; (reverse '((1 2) (3 4) (5 6))) returns ((5 6) (3 4) (1 2)) ;; (reverse "hello") generates and error, because the argument must be a list ;; ;; (mirror 1234) returns 1234 ;; (mirror '( )) returns ( ) ;; (mirror '((1 2) (3 4) (5 6))) returns ((6 5) (4 3) (2 1)) ;; (mirror '((1 2) ((3) ("4" (5 "6")) 7))) returns ((7 (("6" 5) "4") (3)) (2 1)) ;; (define (mirror item) (

6 c) (7 points) Write the all-divisions function, which takes a list of primitives and returns a list of all possible divisions, where a division is any pair of lists whose concatenation is the original list. 6 # kawa:1 # (all-divisions '()) ((() ())) # kawa:2 # (all-divisions '(1 2 3)) ((() (1 2 3)) ((1) (2 3)) ((1 2) (3)) ((1 2 3) ())) # kawa:3 # (all-divisions '( )) ((() ( )) ((1) ( )) ((1 2) ( )) ((1 2 3) ( )) (( ) ( )) (( ) ( )) (( ) ( )) (( ) ( )) (( ) ( )) (( ) ( )) (( ) ( )) (( ) ( )) (( ) ( )) (( ) (14 15)) (( ) (15)) (( ) ())) Your implementation should use a combination of car-cdr recursion, mapping, and anonymous inner functions to get the job done. In particular, you should recursively compute the all-divisions result on the cdr of any non-null list of primitives, and then map over that list to press the car onto the first of the two lists in each division. Feel free to tear this page out, and present your implementation on the next page.

7 7 (more space for Problem 2c) ;;; ;; Returns the list of all the different ways to divide an ordered list ;; into two lists, such that the concatenation of the two lists gives you the ;; original. Check the output format carefully: The result is a list of ;; lists, which themselves contain two lists of primitives. ;; (define (all-divisions sequence) (

8 8 Problem 3: C/Scheme Comparison (7 points) a. (4 points) C and Scheme differ in their treatment of functional arguments. The quicksort we used in Assignment 7 can be seen as the rough equivalent to the low-level qsort provided by ANSI C; both even take a functional argument as a parameter. Give an example of a type of error you can make involving the functional argument to quicksort/qsort that is caught by C at compile-time but not caught by Scheme until runtime. Then give an example of a type of mistake involving the functional argument to quicksort detected by Scheme that is not caught by C's error-checking at all. b. (3 points) Consider the following similar code fragments, one in Scheme, and one in C: (let ((a (<some-expression>))) a = <some-expression>; (+ (func1 a) (func2 a))) return func1(a) + func2(a); You are trying to build a wonder compiler that can transform sequential programs (i.e. authored with no explicit support for concurrency and synchronization) into ones that operate in parallel. To convert the above fragments, you propose dispatching two concurrent threads, one to compute func1 on argument a, the other to compute func2, wait for the two results and then do the addition. Explain why this transformation will be safe and correct for the Scheme code but not always for the C version.

9 9 Problem 4: Java Runtime (7 points) Consider the following hierarchy, which helps to model various items from the Starbucks menu. public abstract class Beverage { public Beverage() { sugar = 10; public void stir() { steam(); sip(); public static void sip() { echo("beverage.sip"); public abstract void steam(); protected int sugar; protected static void echo(string text) { System.out.println(text); public class Latte extends Beverage { public Latte() { drink(); public void drink() { echo("latte.drink " + sugar); public void steam() { echo("latte.steam"); drink(); public class VanillaLatte extends Latte { public VanillaLatte() { sugar = 2; public void stir() { VanillaLatte vl = new VanillaLatte(); vl.sip(); super.stir(); public static void sip() { echo("vanillalatte.sip"); public void drink() { echo("vanillalatte.drink " + sugar); Consider the following method that is declared to take a Beverage reference as a parameter: void starbucks(beverage bev) { bev.sip(); bev.stir(); What are the possible types of objects that bev may be referencing at runtime? For each possibility, trace through a call to the starbucks method and present the output. Write small.

10 10 Problem 5: Java Short Answer (7 points) a. (4 points) You are given the following class definitions: public class A { public void foo(a a) { System.out.println("Hi!"); public class B extends A { public void bar(b b) { System.out.println("Bye!"); Which of the following method calls gives a compiler or a runtime error? Please enter Yes or No in each box in the table. If the line doesn't compile, ignore the corresponding runtime error box. { A a = new A(); B b = new B(); CT Error: (Yes or No) RT Error: (Yes or No) a.foo(b); a.foo((a)b); a.bar(a); b.bar(a); b.bar((b)a); b.foo(b); b. (3 points) Java has several features that make it safer but less efficient than C. Give two examples of such features that provide run-time safety at the cost of run-time efficiency.

11 11 Problem 6: Smart itunes Threading (10 points) You ve been given the following thread-safe, sequential function that downloads a single song from wherever it is that Apple stores its music: bool DownloadSong(const char *songname); DownloadSong returns true if and only if the song was successfully downloaded, and false if there were any problems whatsoever. Writing a function called DownloadSelections that, given an arbitrary long array of song titles, using concurrency and multithreading to download all of the songs. However, instead of spawning a separate thread for each download, spawn just eight threads to share the downloading responsibilities. Each thread is spawned, and then keeps cycling in its own while loop until it detects that all songs have been downloaded. With each iteration of its while loop, it pulls one more song (making sure that no song gets pulled twice) from the list. Don t assume that each of the eight threads downloads just one eighth of all the songs. Implement the download threads to download any song that has yet to be downloaded. Once all songs have been downloaded, each of the threads signals DownloadSelections that it s done, and only when all child threads have signaled DownloadSong can DownloadSelections return with the number of songs that were downloaded successfully. You may assume that RunAllThreads has already been called. Don t use any global variables at all, but instead declare all shared data within DownloadSelections and share their locations with each of the eight threads you spawn. Don t orphan any memory, and don t allow DownloadSelections to return until all songs have been downloaded and all eight child threads have finished. No race conditions, no deadlock. Use this and the next page for your work: int DownloadSelections(const char *songnames[], int numsongs) {

12 (more space for Problem 6) 12