COMPUTER SYSTEMS&MOD MSGT/ET SA AC PKG
3rd Edition
ISBN: 9780134671123
Author: Bryant
Publisher: PEARSON
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Chapter 12, Problem 12.29HW
Program Plan Intro
Deadlocks:
- The deadlock denotes a situation where a collection of threads remains blocked.
- It denotes waiting for a condition that will never be true.
- The progress graph denotes invaluable tool for understanding deadlock.
- If “P” and “V” operations are been ordered incorrectly, the forbidden regions for two semaphores would overlap.
- If some execution trajectory reaches deadlock state “d” then no further progress is possible.
- In this case each thread waits for the other one for a “V” operation that never occurs.
- The overlapping forbidden regions would induce a set of states called “deadlock region”.
- The deadlock is inevitable is trajectory touches a state in deadlock region.
- Once it enters deadlock regions, the trajectories could never leave.
Mutex lock ordering rule:
- A program is deadlock-free if each thread acquires its mutexes in order.
- It releases them in reverse order and given a total ordering for all mutexes.
- The mutex should be achieved in order to ensure deadlock free scheme.
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D. S. Malik, Data Structures Using C++, 2nd Edition, 2010
Methods (Ch6) - Review
1. (The MyRoot method) Below is a manual implementation of the Math.sqrt() method in Java.
There are two methods, method #1 which calculates the square root for positive integers, and
method #2, which calculates the square root of positive doubles (also works for integers).
public class SquareRoot {
public static void main(String[] args) {
}
// implement a loop of your choice here
// Method that calculates the square root of integer variables
public static double myRoot(int number) {
double root;
root=number/2;
double root old;
do {
root old root;
root (root_old+number/root_old)/2;
} while (Math.abs(root_old-root)>1.8E-6);
return root;
}
// Method that calculates the square root of double variables
public static double myRoot(double number) {
double root;
root number/2;
double root_old;
do {
root old root;
root (root_old+number/root_old)/2;
while (Math.abs (root_old-root)>1.0E-6);
return root;
}
}
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Chapter 12 Solutions
COMPUTER SYSTEMS&MOD MSGT/ET SA AC PKG
Ch. 12.1 - Prob. 12.1PPCh. 12.1 - Prob. 12.2PPCh. 12.2 - Practice Problem 12.3 (solution page 1036) In...Ch. 12.2 - Practice Problem 12.4 (solution page 1036) In the...Ch. 12.4 - Prob. 12.5PPCh. 12.4 - Prob. 12.6PPCh. 12.5 - Prob. 12.7PPCh. 12.5 - Prob. 12.8PPCh. 12.5 - Prob. 12.9PPCh. 12.5 - Prob. 12.10PP
Ch. 12.6 - Prob. 12.11PPCh. 12.7 - Prob. 12.12PPCh. 12.7 - Prob. 12.13PPCh. 12.7 - Prob. 12.14PPCh. 12.7 - Prob. 12.15PPCh. 12 - Prob. 12.20HWCh. 12 - Derive a solution to the second readers-writers...Ch. 12 - Prob. 12.22HWCh. 12 - Prob. 12.23HWCh. 12 - Prob. 12.24HWCh. 12 - Prob. 12.25HWCh. 12 - Prob. 12.26HWCh. 12 - Some network programming texts suggest the...Ch. 12 - Prob. 12.28HWCh. 12 - Prob. 12.29HWCh. 12 - Prob. 12.30HWCh. 12 - Implement a version of the standard I/O fgets...Ch. 12 - Prob. 12.32HWCh. 12 - Prob. 12.33HWCh. 12 - Prob. 12.34HWCh. 12 - Prob. 12.35HWCh. 12 - Prob. 12.36HWCh. 12 - Prob. 12.37HWCh. 12 - Prob. 12.38HWCh. 12 - Prob. 12.39HW
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