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quiz with Solution on System Programming | CS 241, Quizzes of Computer Science

University of Illinois - Urbana-ChampaignComputer Science
Prof. Indranil Gupta

Material Type: Quiz; Professor: Gupta; Class: System Programming; Subject: Computer Science; University: University of Illinois - Urbana-Champaign; Term: Spring 2006;

Typology: Quizzes

Pre 2010

Uploaded on 03/10/2009

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Quiz 5
Course: cs241 - System Programming, CS Department
Date: March 5, 2006
Netid: Name: UIN:
Note: Completion of quiz is an individual effort. The quiz takes 10 minutes. The student
gets additional 5 points for taking the quiz. Each question has ONLY ONE ANSWER!!!
1. (1 Point) Given a system where all shared resource types are numbered, if a process is
ordering a resource of type N, it should not be holding a resource of type greater than
or equal to N. This is a method for deadlock
a. Prevention
b. Detection
c. Recovery
2. (1 Point) A deadlock detection algorithm is run
a. Every time a resource is allocated
b. Every time a resource is released
c. As frequently as is decided by the system administrator
3. (1 Point) Let us consider two processes A and B and three resource types R, S, and T.
Let us consider that process A requests resource types R, S, and T, and process B
requests two of the resource types R and S. If A gets assigned resource types R and
T, and B gets assigned resource type S, then
a. A and B will be in deadlock situation
b. There is no deadlock situation between A and B
4. (2 Points) To measure the running time of a function function_to_time by
using POSIX:TMR clocks we need to run the following piece of code (error checking
is omitted):
long timedif;
struct timespec ovalue; nvalue;
clock_gettime(CLOCK_REALTIME, &ovalue);
function_to_time();
clock_gettime(CLOCK_REALTIME, &nvalue);
To compute the running time in microseconds the following formula is correct:
a. timedif = MILLION*(nvalue.tv_sec โ€“ ovalue.tv_sec) +
(nvalue.tv_usec โ€“ ovalue.tv_usec)/1000;
b. timedif = MILLION*(ovalue.tv_sec-nvalue.tv_sec)
+(ovalue.tv_nsec โ€“ nvalue.tv_nsec)
c. timedif= MILLION*(nvalue.tv_sec-ovalue.tv_sec)
+(nvalue.tv_nsec โ€“ ovalue.tv_nsec)/1000
d. timedif = MILLION*(ovalue.tv_sec โ€“ nvalue.tv_sec) +
(ovalue.tv_usec โ€“ nvalue.tv_usec)
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Quiz 5

Course: cs241 - System Programming, CS Department

Date: March 5, 2006

Netid: Name: UIN:

Note: Completion of quiz is an individual effort. The quiz takes 10 minutes. The student gets additional 5 points for taking the quiz. Each question has ONLY ONE ANSWER!!!

  1. (1 Point) Given a system where all shared resource types are numbered, if a process is ordering a resource of type N , it should not be holding a resource of type greater than or equal to N. This is a method for deadlock a. Prevention b. Detection c. Recovery
  2. (1 Point) A deadlock detection algorithm is run a. Every time a resource is allocated b. Every time a resource is released c. As frequently as is decided by the system administrator
  3. (1 Point) Let us consider two processes A and B and three resource types R, S , and T. Let us consider that process A requests resource types R, S , and T , and process B requests two of the resource types R and S. If A gets assigned resource types R and T, and B gets assigned resource type S, then a. A and B will be in deadlock situation b. There is no deadlock situation between A and B
  4. (2 Points) To measure the running time of a function function_to_time by using POSIX:TMR clocks we need to run the following piece of code (error checking is omitted): long timedif; struct timespec ovalue; nvalue;

clock_gettime(CLOCK_REALTIME, &ovalue); function_to_time(); clock_gettime(CLOCK_REALTIME, &nvalue);

To compute the running time in microseconds the following formula is correct: a. timedif = MILLION(nvalue.tv_sec โ€“ ovalue.tv_sec) + (nvalue.tv_usec โ€“ ovalue.tv_usec)/1000; b. timedif = MILLION(ovalue.tv_sec-nvalue.tv_sec) +(ovalue.tv_nsec โ€“ nvalue.tv_nsec) c. timedif= MILLION(nvalue.tv_sec-ovalue.tv_sec) +(nvalue.tv_nsec โ€“ ovalue.tv_nsec)/ d. timedif = MILLION(ovalue.tv_sec โ€“ nvalue.tv_sec) + (ovalue.tv_usec โ€“ nvalue.tv_usec)

  1. (1 Point) Interval timers are useful in programming a. Queue structures b. Process schedulers c. Convey effects
  2. (1 Point) To get into a deadlock situation, we need to have a. Mutual exclusion AND hold and wait situation AND no preemption AND circular wait situation b. Mutual exclusion AND preemption AND spooling everything AND ordering of resources numerically c. No preemption AND hold and wait situation AND no preemption AND ordering of resources numerically
  3. (1 Point) What is the difference when measuring run time of a function_to_time using POSIX:XSI timer functions versus POSIX:TMR timer functions? a. XSI timer is more precise than TMR timer b. TMR timer is more precise than XSI timer c. TMR timer can use virtual time where XSI timer is allowed to use only clock real-time.
  4. (1 Point) What does the following code do (error checking omitted) when we pass sec parameter with value 2.5?

static int functionE(double sec) { timer_t timerid; struct itimerspec tpend;

timer_create(CLOCK_REALTIME, NULL, &timerid); tpend.it_interval.tv_sec = (long) sec; tpend.it_interval.tv_nsec = (sec โ€“ tpend.it_interval.tv_sec)*BILLION; tpend.it_value = tpend.it_interval; return timer_settime(timerid,0,&tpend, NULL); }

a. Set the timer to the current time b. Set the timer to the interval equal to the value stored in CLOCK_REALTIME c. Set the timer to the interval equal to the value of 2. d. Set the timer to the interval equal to the value stored in timerid;