CSSE 332 OPERATING SYSTEMS

Transcription

1 CSSE 332 OPERATING SYSTEMS EXAM 1 MAKEUP SOLUTIONS, SPRING This exam is to be done on paper with limited computer usage. The exam consists of 12 problems on 11 pages for a total of 100 points. INSTRUCTIONS FOR TAKING THE EXAM The exam is open book, open notes, and limited computer usage. You may use your computer only to access your slides, notes, and any document accessible from the class Angel page. Write all of your answers on these pages. You may use the back of a page if you need more space, but indicate on the front when you do so. When you complete the exam, turn it in to the exam proctor. Use of any resource other than those mentioned above before turning in the exam will be considered academic dishonesty. Please begin by putting your name on every page of the exam. You are encouraged to skim the entire exam before answering any question. In order to receive full credit for your solutions, you are advised to write legibly so your grader can read your solutions. Prob. Points Score P1 15 P2 10 P3 10 P4 6 P5 4 P6 14 P7 6 P8 6 P9 8 P10 11 P11 6 P12 4 Total 100 Page 1 of 9

2 C CODE 1. (15 points) What is the result of compiling and executing each of the following C programs? a. (7 points) struct.c #include <stdio.h> struct numbers{ double first; double second; ; typedef struct numbers numeros; double c(numeros those) { if (those.first < those.second) { those.first = those.second - those.first; else { those.second = those.first - those.second; printf( First is %1.2lf; Second is %1.2lf\n, those.first, those.second); return those.first + those.second; double d(numeros numbs) { numeros reals; reals.first = numbs.second; reals.second = 2 * numbs.first; return c(reals) - c(numbs); int main() { numeros these = {5.5, 7.5; numeros some = {7.5, 5.5; printf("%1.2f\n", d(these)); printf("%1.2f\n", d(some)); return 0; Output: First is 3.50; Second is First is 2.00; Second is First is 9.50; Second is First is 7.50; Second is Page 2 of 9

3 b. (8 points) modify.c void modifyme(int x, int y, int *px, int *py) { x = x + y; *px += y; py = px; int main() { int a, b, c, d; a = 10; b = 20; c = 30; d = 40; modifyme(a, b, &c, &d); printf("%d, %d, %d, %d\n", a, b, c, d); return 0; Output: 10, 20, 50, (10 points) Given the declarations below, answer the questions that follow. /* Initial size of arraylist */ #define INIT_SIZE 4 typedef struct { //Pointer to dynamically allocated array to hold elements. int* array; // The number of elements in arraylist. int size; // Maximum size of arraylist before it can be resized. */ int capacity; arraylist; a. (2 points) Write code that declares a variable of type arraylist. arraylist myarray; b. (4 points) Write code to initialize the variable declared in part (a). Every element of the struct must be initialized using only the information in the declarations given above. Consider using the malloc function as needed. myarray.size = 0; myarray.array = (int *) malloc(init_size * sizeof(int)); myarray.capacity = INIT_SIZE; Page 3 of 9

4 c. (4 points) Write code to double the capacity of the arraylist from part (b). Use appropriate memory management functions to actually double the length of the array. int nsize = myarray.capacity * 2; myarray.array = (int *)realloc(myarray.array, \ nsize*sizeof(int)); myarray.capacity = nsize; C PROGRAM COMPILATION AND LINUX ENVIRONMENT 3. (10 points) Given that the executable file project depends on the files main.c, io.c, and io.h, as illustrated in the Makefile below, answer the following questions. CC = gcc CFLAGS = -g -Wall PROG = project HDRS = io.h SRCS = main.c io.c OBJS = $(SRCS:.c=.o) $(PROG): $(OBJS) $(CC) $(OBJS) -o $(PROG) main.o: main.c io.h // Group 1 starts here // Group 2 starts here // Group 3 starts here // Group 4 starts here // Group 5 starts here io.o: io.c io.h a. (6 points) Explain what each group of makefile rules does. Group 1: Variable declaration, name of the default compiler and list of options to pass to the compiler. Group 2: Variable declaration, name of executable program and list of header files and source files. Group 3: Variable declaration, give object files same name as source files, except extensions. Page 4 of 9

5 Group 4: Build the executable from object files. Group 5: Rules for compiling the source files b. (2 points) Why are there no compilation rules for the dependencies listed in Group 5? Because by default they compile each <file>.c to <file>.o c. (2 points) Suppose I modify io.c and run make. Which of the target files will be modified? io.o and project 4. (10 points) Assume you have checked out project Test from your svn repository to your local workspace and have since created the files main.c, io.c, io.h, driver.c, and Makefile and put them in your local workspace. Write complete svn commands to do the following. Note: every commit command must be accompanied with an appropriate commit message. a. (2 points) Ensure that your local workspace has the latest version of the files in your svn repository. svn up. b. (5 points) Ensure that your svn repository has exactly the same version of every file in you local workspace. svn add * svn commit m Committed my changes 5. (4 points ) Answer the following svn-related questions: a. (2 points) You and your partner have access to a shared svn repository. Consider a situation where you and your partner happen to edit your respective local copies of a certain file that you retrieved from the shared svn repository simultaneously. Your partner commits his/her version of the file to the shared svn repository first and sometime later you attempt to commit to the svn repository your version of the same file. What do you think will happen? Svn might complain that your local copy is out of date (causing conflicts). It should ask you to update your local copy before committing. The update should try to merge Page 5 of 9

6 changes and resolve any conflicts. If merging fails, you will have to merge and resolve conflicts manually before committing. b. (2 points) Assuming that you have been committing your changes to your svn repository regularly, what would you do to retrieve a specific version of a file? Note: If you don t know the specific commands, describe the process. You need to use svn log to find the revision number that interests you. You then do an svn update r <revision_number> to update the file to that revision number. STALLINGS CHAPTERS 1, 2, 3, AND 4 6. (14 points) Consider a hypothetical microprocessor generating a 32-bit address (for example, assume the program counter and the address registers are 32 bits wide) and having a 16-bit data bus. Assume that memory is byte addressable. Where appropriate, give your answer in appropriate units of Bytes. a. (3 points) What is the maximum address space that the processor can access directly if it is connected to a 36-bit memory? 2 32 Bytes = 4GB b. (3 points) What is the size of the memory? 2 36 Bytes = 64 GB c. (8 points) Assuming each process requires a minimum of 2MB and a maximum of 256GB. i. What is the smallest number of processes that can fit in memory at the same time? O processes since the maximum memory requirements of a process cannot be satisfied with limited memory available. ii. What is the largest number of processes that can fit in memory at the same time? 2 36 Bytes / 2 21 Bytes = 2 15 processes or processes 7. (6 points) Describe the round-robin scheduling algorithm. The round-robin scheduling algorithm dispatches the process at the front of the ready queue and allows it to run for a certain fixed time slice. After its time slice expires, the Page 6 of 9

7 process is put on the back of the queue (providing it has not terminated and its state does not change otherwise) and the next process at the front of the queue is dispatched. 8. (6 points) Contrast the scheduling policies you might use when trying to optimize a time-sharing system with those you would use to optimize a multiprogrammed batch system. Time sharing systems favor scheduling policies that minimize response time or turnaround time. Examples of such policies include preemptive policies and time-slicing policies. Multiprogrammed batch systems care mostly about optimizing throughput and processor usage. This means that simple scheduling policies that devote more processing time to processes are favored. Such policies minimize process switches. 9. (8 points) Assume that a particular operating system uses the five-state process model, and consider the sequence of events given below. i. Process P1 is created, admitted and dispatched. ii. Process P2 is created and admitted. iii. Process P3 is created and admitted. iv. Process P4 is created and admitted. v. The running process issues an I/O request. vi. A process is dispatched. vii. The running process times out. a. (3 points) At the end of the sequence, what is executing? Hint: it s not P3. At the end of the sequence, no processes are in the running state. Thus, the kernel is executing. It would be acceptable to answer that the dispatcher is running, although strictly speaking the timer interrupt service routine must invoke the dispatcher, so either the ISR or the dispatcher could be executing depending on the exact instant considered. Furthermore, depending on the relationship between the operating system and user processes (see Figure 3.15), other kernel code could also execute before the next process is dispatched. b. (3 points) Which process is at the head of the ready queue? P3 c. (2 points) Which process is at the tail of the ready queue (assume that the I/O request has not completed)? P2 Page 7 of 9

8 10. (13 points) Answer the following short answer questions on process description and control. a. (3 points) The operating system uses control tables to keep track of resources that it manages. Identify three such tables. Memory tables, I/O tables, file tables, primary process table b. (4 points) Describe the difference between blocking a process and suspending a process. Blocking a process keeps the process in memory but moves it to the blocked queue. Suspending a process moves it to secondary storage. c. (2 points) Explain why a process switch is always slower than a mode switch? A process switch requires a mode switch to kernel mode, execution of the dispatcher, and another mode switch back to user mode. Thus, process switches necessarily take more than twice as long as mode switches. In practice, they take at least one order of magnitude longer. d. (2 points) What does it mean to preempt a process? To preempt a process means to forcibly retrieve the processor or some other resource from the process so another process can execute. 11. (6 points) You have written, compiled, and executed the following C program: #include <stdio.h> #include <stdlib.h> int main(){ int pid; pid = fork(); printf("%d %d\n", pid, getpid()); Assuming the fork() system call succeeded, specify two possible outputs of the program s execution. Also assume that the parent process process id is 2536 and that of any child created will be greater than a. (3 points) Page 8 of 9

9 In this case fork() returns in the child before it returns in the parent b. (3 points) In this case fork() returns in the parent before it returns in the child 12. (4 points) Examine the following code segment and answer the questions below: #include <stdio.h> #include <stdlib.h> int main (){ int pid; for(;;){ pid = fork(); a. (3 points) Assuming the program compiled and you are able to execute the program, identify the problem with the program. Both the parent and its descendents keep spawning processes until the system either runs out of process IDs, memory, or crashes. This problem is called fork bomb. Wikipedia has a good discussion of this. b. (1 point) How could you remedy this situation? As a user, do not execute the program. If you do, you may have to restart your computer, since the system may not have a pid left to allow you to create a process to terminate the program. As a system admin or OS developer, limit the number of processes a user can spawn. That number should be smaller than the number of processes the system can support at once. Page 9 of 9