Final answer:
The student needs to implement three methods in the ArrayListMethods class: print, condense, and duplicate. Each method manipulates an ArrayList in different ways, such as printing all elements, condensing the list to half the size by combining elements, and duplicating each element in the original list.
Explanation:
To complete the ArrayListMethods class, start by importing the java.util.ArrayList package. Then, define each method with the appropriate logic. For example, your print method should iterate through the ArrayList, printing each element.
ArrayListMethods.java
import java.util.ArrayList;
public class ArrayListMethods {
public static void print(ArrayList list) {
for (String element : list) {
System.out.println(element);
}
}
public static void condense(ArrayList list) {
for (int i = 0; i < list.size() - 1; i++) {
list.set(i, list.get(i) + list.get(i + 1));
list.remove(i + 1);
}
}
public static void duplicate(ArrayList list) {
for (int i = 0; i < list.size(); i += 2) {
list.add(i + 1, list.get(i));
}
}
}
The condense method reduces the size of the ArrayList by half by merging adjacent elements, while the duplicate method doubles the size of the ArrayList by duplicating each element.
Which layer of the OSI reference model is responsible for ensuring flow control so that the destination station does not receive more packets that it can process at any given time? Group of answer choices
Answer: The Transport Layer of the OSI reference model is responsible for ensuring flow control so that destination station does not receive more packets that it can process at any given time.
Explanation: This is because;
The transport layer is the fourth layer in the OSI layered architecture which builds on the network layer to provide data transport that moves from a process on a source machine to a process on a destination machine. It is hosted using single or multiple networks, and very responsible for reliable data delivery ensuring packets are delivered in sequence, error-free and with little or no duplication or losses.
Since Transport layer helps one to control the reliability of a link through flow control, error control, and segmentation or desegmentation, It determines how much data should be sent where and at what rate.
The transport layer also offers an acknowledgment of the successful data transmission and sends the next data in case no errors occurred. TCP (Transmission Control Protocol )is the best-known example of the transport layer. Transport layers also retransmit messages if they arrive with errors.
Digital subscriber lines: are very-high-speed data lines typically leased from long-distance telephone companies. are assigned to every computer on the Internet. operate over existing telephone lines to carry voice, data, and video. have up to twenty-four 64-Kbps channels. operate over coaxial cable lines to deliver Internet access.
Answer: Operate over existing telephone lines to carry voice, data, and video.
Explanation:
Digital subscriber line is a means of transferring high bandwidth data over a telephone line. Such data could be a voice call, graphics or video conferencing. DSL uses a user's existing land lines in a subscriber's home, allowing users to talk on a telephone line while also being connected to the Internet. In most cases, the DSL speed is a function of the distance between a user and a central station. The closer the station, the better its connectivity.
Write a loop that sets each array element to the sum of itself and the next element, except for the last element which stays the same. Be careful not to index beyond the last element. Ex: Initial scores: 10, 20, 30, 40 Scores after the loop: 30, 50, 70, 40 The first element is 30 or 10 20, the second element is 50 or 20 30, and the third element is 70 or 30 40. The last element remains the same.
Answer:
The solution code is written in Python
numList = [10, 20, 30, 40] for i in range(0, len(numList) - 1): numList[i] = numList[i] + numList[i + 1] print(numList)Explanation:
Firstly, create a sample number list, numList (Line 1)
Create a for-loop that will traverse through the array element from 0 till the second last of the element (len(numList) - 1) (Line 3)
Set the current element, numList[i], to the sum of the current element, numList[i] and the next element, numList[i+1]
Print the modified numList (Line 6) and we can see the output as follows:
[30, 50, 70, 40]
Suppose there are two ISPs providing WiFi service in a café. Each ISP operates its own AP and has its own IP address block. If by chance both ISPs configure their APs to operate over the same channel, e.g channel 5, how will users who attempt to connect to either of the APs be affected?
Users trying to connect to APs set on the same Wi-Fi channel will likely face interference, resulting in connection instability and reduced performance. Such channel overlap can mimic a Denial of Service, slowing down or disrupting network access.
If two Internet Service Providers (ISPs) configure their Access Points (APs) to operate on the same Wi-Fi channel, such as channel 5, the users trying to connect to either AP may experience interference and degradation in Wi-Fi quality. This interference can cause a Denial of Service (DoS) effect, where users may face difficulty establishing a stable connection due to heavy traffic on the same channel. Signals may overlap, which leads to signal contention, slower speeds, increased latency, and potentially dropped connections. It is similar to when a microwave oven causes interference with a Wi-Fi system, as both are emitting signals in a similar frequency range.
If the channel congestion is severe, it can feel akin to a DoS attack in which the AP is overwhelmed, preventing legitimate users from gaining or maintaining a smooth connection. To mitigate such issues, APs should be set to operate on different channels, or technologies like connection manager software and mobile Virtual Private Networks (VPNs) can be leveraged for a more stable experience.
You can view the existing Access Control Lists for a set of folders on a Windows system by right-clicking the folder you want to view, selecting Properties, and clicking the:
Answer:
And clicking the security tab option.
Explanation:
Lets explain what an object's ACL is. I will use an example to best explain this. Let's suppose that user Bob would want to access a folder in a Windows environment. What supposedly will happen is that Windows will need to determine whether Bob has rights to access the folder or not. In order to do this, an ACE with the security identity of John will be created. These ACEs are the ones that grant John access to the folder and the ACLs of this particular folder that John is trying to access is a list of permissions of everyone who is allowed to access this folder. What this folder will do is the to compare the security identity of John with the folders ACL and determine whether John has Full control of the folder or not.
By right clicking the folder and selecting the security tab, John will be in a position to see a list of the permissions (ACLs) granted to him by the folder.
this IDS defeating techniques works by splitting a datagram or packet into multiple fragments and the IDS will not spot the true nature of the fully assembled datagram. the datagram is not reassembled until it reaches its final destination. It would be a processor-intensive task for an IDS to reassemble all fragments itself, and on a busy system the packet will slip through the IDS onto the network. what is this technique called?
A. IP routing or packet dropping
B. IP splicing or packet reassesmbly
C. IDS spoofing or session assembly
D. IP fragmentation or Session splicing
Answer:
D. IP Fragmentation or Session Splicing
Explanation:
The basic premise behind session splicing, or IP Fragmentation, is to deliver the payload over multiple packets thus defeating simple pattern matching without session reconstruction. This payload can be delivered in many different manners and even spread out over a long period of time. Currently, Whisker and Nessus have session splicing capabilities, and other tools exist in the wild.
Which of the following statements regarding the current electronic waste scenario is true? Electronic waste increases with the rise of living standards worldwide. The content of gold in a pound of electronic waste is lesser than that in a pound of mined ore. The process of separating densely packed materials inside tech products to effectively harvest the value in e-waste is skill intensive. Sending e-waste abroad can be much more expensive than dealing with it at home. E-waste trade is mostly transparent, and stringent guidelines ensure that all e-waste is accounted for.
Answer:
Electronic waste increases with the rise of living standards worldwide.
Explanation:
Electronic waste are electronic product nearing the end of their usefulness or discard ones, it include computers, phones ,oven, scanners, keyboard,circuit board,clock, lamp etc
Their amount increases every year with the rise of living standards worldwide which constitute to million of tonnes in circulation or these waste end up in landfill which at time contaminate the soil and water by it lead and cadmium content .The amount of electronic waste is expected to increase to about 52.2 million metric tonnes by 2021
Final answer:
True statements regarding the e-waste scenario include the increase of e-waste with higher living standards, the skill-intensive process of recycling, and the high content of gold in e-waste compared to mined ore. However, the e-waste trade is not transparent and is often improperly managed.
Explanation:
Among the current statements about the electronic waste (e-waste) scenario, several are true. The statement that e-waste increases with the rise of living standards worldwide is true. As technological advancements continue and consumerism grows, the amount of e-waste generated increases.
Also true is the fact that the process of separating densely packed materials inside tech products to effectively harvest the value in e-waste is skill-intensive, often leading to unsafe and unregulated labor practices in developing countries. It is indeed accurate that the content of gold in a pound of electronic waste is often greater than that in a pound of mined ore, making it a highly valuable resource.
However, contrary to one of the statements, the e-waste trade is not mostly transparent, and stringent guidelines do not ensure that all e-waste is accounted for; in fact, there is a significant amount of e-waste that is illegally traded or improperly handled.
Bob received a message from Alice which she signed using a digital signature. Which key does Bob use to verify the signature?Group of answer choicesAlice's private keyBob's public keyAlice's public keyBob's private key
Answer:
Alice's public key
Explanation:A Public key is a key that can be used for verifying digital signatures generated using a corresponding private key which must have been sent to the user by the owner of the digital signature.
Public keys are made available to everyone required and they made up of long random numbers.
A digital signature signed with a person's private key can only be verified using the person's private key.
The variable planet_distances is associated with a dictionary that maps planet names to planetary distances from the sun. Write a statement that deletes the entry for the planet name "Pluto".
Answer:
Check the attached image
Explanation:
Check the attached image
Create an array w with values 0, 0.1, 0.2, . . . , 3. Write out w[:], w[:-2], w[::5], w[2:-2:6]. Convince yourself in each case that you understand which elements of the array that are printed.
Answer:
w = [i/10.0 for i in range(0,31,1)]
w[:] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3.0]
w[:-2] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8]
w[::5]= [0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0]
w[2:-2:6] = [0.2, 0.8, 1.4, 2.0, 2.6]
Explanation:
List slicing (as it is called in python programming language ) is the creation of list by defining start, stop, and step parameters.
w = [i/10.0 for i in range(0,31,1)]
The line of code above create the list w with values 0, 0.1, 0.2, . . . , 3.
w[:] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3.0]
since start, stop, and step parameters are not defined, the output returns all the list elements.
w[:-2] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8]
w[:-2] since stop is -2, the output returns all the list elements from the beginning to just before the second to the last element.
w[::5]= [0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0]
w[::5] since step is -2, the output returns the list elements at index 0, 5, 10, 15, 20, 25,30
w[2:-2:6] = [0.2, 0.8, 1.4, 2.0, 2.6]
the output returns the list elements from the element at index 2 to just before the second to the last element, using step size of 6.
You are to design class called Employee whose members are as given below:
Data members:
char *name
long int ID
A constructor:
// initialize data members to the values passed to the constructor
Methods:
get Person (name, id) //allows user to set information for each person
A function called Print () that prints the data attributes of the class.
Show the working of class in a driver program (one with main).
Answer:
//The Employee Class
public class Employee {
char name;
long ID;
//The constructor
public Employee(char name, long ID) {
this.name = name;
this.ID = ID;
}
//Method Get Person
public void getPerson (char newName, long newId){
this.ID = newName;
this.ID = newId;
}
//Method Print
public void print(){
System.out.println("The class attributes are: EmpName "+name+" EmpId "+ID);
}
}
The working of the class is shown below in another class EmployeeTest
Explanation:
public class EmployeeTest {
public static void main(String[] args) {
Employee employee1 = new Employee('a', 121);
Employee employee2 = new Employee('b', 122);
Employee employee3 = new Employee('c', 123);
employee1.print();
employee2.print();
employee3.print();
}
}
In the EmployeeTest class, Three objects of the Employee class are created.
The method print() is then called on each instance of the class.
. Identify an emerging crime issue in your community using data available from sources such as local newspapers, online police reporting, and so forth. Frame the situation, and then identify the restraining and driving forces that may be impacting the issue. 2. Using your force field analysis, develop a cause and effect diagram for the situation.
A crime problem in my community is related to cell phone theft. According to the local newspaper, it is estimated that in my city about 10 cell phones are stolen per week. Still according to the local newspaper, most of these robberies occur in the city center and in the periphery, with women being the biggest victims.
Although the police have shown themselves to be a restraining force on this type of crime, few arrests have been made successfully, mainly for the negligence of the victims in providing a complaints.
The main driving force behind this crime is drug trafficking. Most burglars steal cell phones to sell them and have money to buy drugs. This is totally related to the government's neglect to promote quality education in the city, allowing several children and young people to stay on the street and run the risk of becoming involved in the traffic.
A cause and effect diagram for this situation is:
Irresponsible government ---> poor quality education ---> children and adolescents on the streets ---> involvement in drug trafficking ---> theft of cell phones ----> frightened population ---> lack of complaints ----> criminals on the street.
In the sub-module on relations, we discussed total orders. Total orders allow you to sort the elements in a list. Why is sorting such an important operation in computing?
Answer:
The correct answer to the following question will be Option "b" and "d".
Explanation:
The given question is incomplete, options are missing. The complete question is :
(a) A ordered program runs more efficiently.
(b) Queries or claims move faster.
(c) There's less risk the data will be lost.
(d) Although printing out a text is readable.
Now,
Sorting involves organizing the information or data in the sequence to have assented or descending. It's any process that involves organizing the data in such a logical addition to being able to comprehend, understand, interpret or imagine.
Usually, data is ordered in either increasing or decreasing direction which is based on objective numbers or counts but may also be categorized based on the descriptions of the component standards.
So, it's the right answer.
Print "Censored" if userInput contains the word "darn", else print userInput. End with newline. Ex: If userInput is "That darn cat.", then output is:CensoredEx: If userInput is "Dang, that was scary!", then output is:Dang, that was scary!Note: If the submitted code has an out-of-range access, the system will stop running the code after a few seconds, and report "Program end never reached." The system doesn't print the test case that caused the reported message.#include #include using namespace std;int main() {string userInput;getline(cin, userInput);int isPresent = userInput.find("darn");if (isPresent > 0){cout << "Censored" << endl; /* Your solution goes here */return 0;}
Answer:
#include <string>
#include <iostream>
using namespace std;
int main() {
string userInput;
getline(cin, userInput);
// Here, an integer variable is declared to find that the user entered string consist of word darn or not
int isPresent = userInput.find("darn");
if (isPresent > 0){
cout << "Censored" << endl;
// Solution starts here
else
{
cout << userInput << endl;
}
// End of solution
return 0;
}
// End of Program
The proposed solution added an else statement to the code
This will enable the program to print the userInput if userInput doesn't contain the word darn
A company has a legacy application using a proprietary file system and plans to migrate the application to AWS. Which storage service should the company
use?
A. Amazon DynamoDB
B. Amazon S3
C. Amazon EBS
D. Amazon EFS
Final answer:
For a legacy application using a proprietary file system migrating to AWS, Amazon EFS (Elastic File System) is the best storage service to use. It supports NFS protocol, making it suitable for applications needing a traditional file system interface without requiring alterations to their existing file system usage.
Explanation:
If a company has a legacy application using a proprietary file system and plans to migrate the application to AWS, the best storage service to use would be Amazon EFS (Elastic File System). Amazon EFS offers a scalable file storage solution for use with AWS Cloud services and on-premises resources. It's built to provide easy, scalable, and reliable storage for applications that require a file system interface and file system semantics.
Amazon EFS is highly suitable for legacy applications being migrated to AWS because it supports NFS (Network File System) protocol, allowing those applications to work without requiring any alterations to the file system usage. Unlike options like Amazon DynamoDB, which is a NoSQL database service for applications that need consistent, single-digit millisecond latency at any scale, or Amazon S3, best for object storage with durability and scalability, EFS provides a more traditional file system interface and file system semantics, making it a more direct match for legacy applications. Lastly, while Amazon EBS (Elastic Block Store) provides block-level storage volumes for use with Amazon EC2 instances, it's not designed to be directly accessed by applications as a file system.
Which of the following statement is true for Service Request Floods A. An attacker or group of zombies attempts to exhaust server resources by setting up and tearing down TCP connections B. It attacks the servers with a high rate of connections from a valid source C. It initiates a request for a single connection
Answer:
The answer is "Option A and Option B"
Explanation:
This is a type of attack, which is mainly used to bring down a network or service by flooding large amounts of traffic. It is a high-rate server from legitimate sources, where an attacker or group of zombies is attempting to drain server resources by creating and disconnecting the TCP link, and wrong choices can be described as follows:
In option C, It can't initiate a single request because when the servers were overloaded with legitimate source links, the hacker may then set up and uninstall TCP links.
In last week's meeting we discussed long and short term costs associated with build an buy scenarios using a house as an example. Take the same concepts and apply them to a software development project scenario. ie building a custom application for the business house versus buying that application and implementing it. Respond here and list out the criteria you would review to consider the decision.....labor costs long and short term, support costs etc.
Answer:
Custom software designs a software package that is targeted to a particular user community and that meets an organization's specific needs. A lot of these things must be taken into account whenever making a "buy vs. create" decision for a custom software.
Purchasing a wrong program may hinder the process for your business while trying to build one can be expensive and time consuming. The study of these two methods should take into account labor costs, long-term and brief-term costs, and infrastructure costs.
The most popular purpose an organization creates or gets a custom product is that it's special to their organization and if the software is designed effectively it will improve the business' productivity and create its own competitiveness edge.
Moreover, creating a custom software requires a great upfront cost and it also takes a long time to build a proper one.
Labor costs for developing a customized product are often greater than purchasing off-the-shelf solution, as the company has to employ a software developer and build an IT team to create and manage the right software.
There would be maintenance costs in the long run but it wouldn't be as enormous as it was in the building and the process of creation. But, the more significant than cost, is the long-term benefit it brings to the business.
A specific application will improve the workflow of the company, allow the company to retain space with the rate and volume expansion, all of which would help bring financial benefits and distinguish the business from other competitors.
Create a program to compute the fee for parking in a garage for a number of hours. The program should: 1. Prompt the user for how many hours parked 2. Calculate the fee based on the following: a. $2.50/hour b. minimum fee is $6.00 c. maximum fee is $20.00 3. Print the result python
Answer:
The ans will be given in the python script below. A picture of the answer is also attached
Explanation:
print("Welcome To Garage Parking Fee Calculator")
hours = float(input("Type the number of hours parked : "))
#fee per hour
rate = 2.40
#multiply rate per hour by the number of hours inputted
price = rate * hours
if price < 6:
price = 6
if price > 20:
price = 20
print("Parking fee is: $", +price)
Heather wants a transition effect applied to the links in the gameLinks list in which a gradient-colored bar gradually expands under each link during the hover event. To create this effect, you will use the after pseudo-element and the content property to insert the bar. Create a style rule for the nav#gameLinks a::after selector that: a) places an empty text string as the value of the content property, b) places the content with absolute positioning with a top value of 100% and a left value of 0 pixels, c) sets the width to 0% and the height to 8 pixels, d) changes the background to a linear gradient that moves to right from the color value rgb(237, 243, 71) to rgb(188, 74, 0), e) sets the border radius to 4 pixels, and f) hides the bar by setting the opacity to 0.
Answer:
Hi there Zelenky! This is a good question to practice style sheets and effects. Please find my answer below.
Explanation:
Below CSS contains the code to answer all parts of the question.
nav#gameLinks a::after {
content: ‘’;
top: 100%;
left: 0px;
width: 0%;
height: 8px;
position: absolute;
background: -webkit-gradient(linear, right, left, from(rgb(237, 243, 71)), to(rgb(188, 74, 0));
border-radius: 4px;
opacity: 0;
}
In this assignment, you will write a complete C program that will act as a simplecommand-line interpreter (i.e., a shell) for the Linux kernel. In writing your shell, you areexpected to use the fork-exec-wait model discussed in class. In particular, you are toimplement the following:• Loop continuously until the user enters quit, which exits your shell.• Inside the loop, you will print your "minor5" prompt and read in the user’scommand, which may consist of a Linux command with 0 or more options andarguments supported by the command. You are expected to read in and processonly 1 command at a time with no pipelining or redirection.• In a child process, you are to execute the command as given, including alloptions and arguments given. If the command is not valid, rather than display an"exec failed" message as shown in class examples, you will simply print outthe command itself with "command not found" as shown in the SAMPLEOUTPUT and then exit the child process. The parent process should wait for thechild process to finish.If you have any questions about this, please contact your instructor, TAs, or IAsassigned to this course to ensure you understand these directions.SAMPLE OUTPUT (user input shown in bold):$ ./a.outminor5> lsa.out grades.txt rec01.txt testdir phone.txt route.txt who.txtdu.txt rec01.c file1 rec01sol.c minor5.cminor5> ls -a -l -ttotal 144-rwx------ 1 cat0299 cat0299 7835 Oct 14 17:39 a.outdrwx------ 4 cat0299 cat0299 4096 Oct 14 17:39 .-rw------- 1 cat0299 cat0299 2665 Oct 14 17:39 minor5.c-rw------- 1 cat0299 cat0299 33 Oct 5 03:30 du.txt-rw------- 1 cat0299 cat0299 33 Oct 5 01:28 file1-rw------- 1 cat0299 cat0299 333 Oct 5 01:02 route.txt
Answer:
/ to access the input scanf()
// and output printf()
#include <stdio.h>
// to access the functions like
// pipe(), fork(), execvp(), dup2()
#include <unistd.h>
// to access the string functions like
// strtok()
#include <string.h>
// to access function wait()
#include <sys/wait.h>
#include <stdlib.h>
int main()
{
// declare a variable to hold the process id
pid_t p_id;
// declare a variable to hold the index value
int array_index;
// declare the string to hold the user input
// as command
char userIn_Command[128];
// use a continuous loop
while (1)
{
// display the prompt for the user
printf("minor5> ");
// read the input from the user
scanf("%[^\n]", userIn_Command);
// check the condition that whether the user
// inputs a command called "quit"
// If the user inputs quit command then exit from
// the script
if (strcmp(userIn_Command, "quit") == 0)
{
printf("\n");
break;
}
// if there are any usage of pipelining or redirection
// display the error message and exit from the script
if (strchr(userIn_Command, '|') != NULL || strchr(userIn_Command, '>') != NULL ||
strchr(userIn_Command, '<') != NULL)
{
printf("Error: Cannot perform the multiple command operations or directions\n");
break;
}
// declare the variables to hold the process
int p_pids[2];
// create the system call to pipe() from kernal
// and display the error message
if (pipe(p_pids) < 0)
{
printf("Error: Pipe creation failed!\n");
}
// create the child process
p_id = fork();
// condition to check whether the fork() is
// created. If so, display an error message
if (p_pids < 0)
{
printf("Error: fork() failed!\n");
break;
}
// if the child process is created
if (p_id == 0)
{
// close pipe in child process
close(p_pids[0]);
// create duplicate of the standard output
dup2(p_pids[1], STDOUT_FILENO);
// close the pipe in child process
close(p_pids[1]);
// declare a variable to store path
// to execute the command
char *command_path;
// declare an array of string to hold the options
char * args[32];
// tokenize the command by using the delimiter at " "(single space)
char *cmd_token = strtok(userIn_Command, " ");
// store the token value
command_path = cmd_token;
args[0] = cmd_token;
array_index = 1;
// loop until all the options in the command are
// tokenized
while (1)
{
// get the next token
cmd_token = strtok(NULL, " ");
// condition to check whether the token is null
// or not
if (cmd_token == NULL)
{
break;
}
// store the token if it is not null
args[array_index] = cmd_token;
// increment the index
array_index++;
}
// last parameter to the command should be NULL */
args[array_index] = NULL;
/* calling exec function with command path and parameters */
if (strcmp(args[0], "cd") == 0 || strcmp(args[0], "history") == 0 ||
strcmp(args[0], "exit") == 0)
{
printf("%s: Command not found\n", args[0]);
break;
}
if (execvp(command_path, args) < 0 )
{
printf("%s: Command not found\n", args[0]);
break;
}
}
else
{
/* closing writing end of pipe in parent process */
close(p_pids[1]);
/* reading ouput written to pipe in child process and
* writing to console */
while (1)
{
char output[1024];
int n = read(p_pids[0], output, 1024);
if (n <= 0)
{
break;
}
output[n] = '\0';
printf("%s", output);
}
/* closing read end of pipe1 */
close(p_pids[0]);
/* waiting until child process complete its execution */
wait(NULL);
}
/* skipping newline character read while scanf() */
getchar();
}
exit(0);
}
Explanation:
int replace_text (Document *doc, const char *target, const char *replacement) The function will replace the text target with the text replacement everywhere it appears in the docu- ment. You can assume the replacement will not generate a line that exceeds the maximum line length; also you can assume the target will not be the empty string The function will return FAILURE if doc, target or replacement are NULL; otherwise the function will return SUCCESS. #de fine DOCUMENT H #de fine MAX PARAGRAPH LINES 20 #de fine MAX PARAGRAPHS 15 #de fine MAX STR SIZE 80 #de fine HIGHLIGHT START STR "[" #define HIGHLIGHT END STR "ן" #de fine SUCCESS #de fine FAILURE - typedef struct { int number of 1ines: char lines[MAX PARAGRAPH LINES 1 [MAX STR SIZE + 11 \ Paragraph; typedef struct { char name [ MAX STR SIZE+ 11 int number of paragraphs; Paragraph paragraphs [MAX PARAGRAPHS]; } Document;
Answer:
int replace(Document *doc, const char *target, const char *replacement){
int i, j, k;
int beginning;
char temp[MAX_STR_SIZE + 1] ;
char *beginning, *end, *occ, *line;
if(doc == NULL || target == NULL || replacement == NULL)
return FAILURE;
for(i = 0; i < doc->total_paragraphs; i++){
for(j = 0; j < doc->paragraphs[i]->total_lines; j++){
line = doc->paragraphs[i]->lines[j];
beginning = line;
end = beginning + strlen(line);
strcpy(temp, "");
while(beginning < end && (occ = strstr(beginning, target))!= NULL){
len = occ - beginning;
strncat(temp, beginning, len);
strcat(temp, replacement);
beginning = occ + strlen(target);
}
strcat(temp, beginning);
strcpy(doc->paragraphs[i]->lines[j], temp);
}
}
return SUCCESS;
}
Explanation:
Loop through total paragraphs and total lines.Store the beginning and ending of paragraph in specific variables.Copy the remainging chars .Finally return SUCCESS.The inverse of a map is a new map where the values of the original map become the keys of the new map and the keys become the values. For example, given the map {1:2, 3:4}, the inverse is the map {2:1, 4:3}. Given the map, d, create the inverse of d. Associate the new map with the variable inverse. You may assume that there are no duplicate values in d (that is no two keys in d map to the same value).
Answer:
The inverse of d would be:
inverse = { }
for key, value in d.items ( ) :
inverse [value] =key|
Answer:
The inverse of d would be:
inverse = { }
for key, value in d.items ( ) :
inverse [value] =key|
a. Carly’s Catering provides meals for parties and special events. In Chapter 3, you created an Event class for the company. The Event class contains two public final static fields that hold the price per guest ($35) and the cutoff value for a large event (50 guests), and three private fields that hold an event number, number of guests for the event, and the price. It also contains two public set methods and three public get methods.
Now, modify the Event class to contain two overloaded constructors
∎ One constructor accepts an event number and number of guests as parameters. Pass these values to the setEventNumber() and setGuests() methods, respectively. The setGuests() method will automatically calculate the event price.
∎ The other constructor is a default constructor that passes "A000" and 0 to the two-parameter constructor.
Save the file as Event.java
b. In Chapter 3, you also created an EventDemo class to demonstrate using two Event objects. Now, modify that class to instantiate two Event objects, and include the following new methods in the class:
∎ Instantiate one object to retain the constructor default values
∎ Accept user data for the event number and guests fields, and use this data set to instantiate the second object. Display all the details for both objects.
Save the file as EventDemo.java.
Here's the modified Event class with two overloaded constructors and the EventDemo class with the required methods:
`java
// Event.java
package event;
public class Event {
public static final int CUTOFF_VALUE = 50;
public static final int PRICE_PER_GUEST = 35;
private final int eventNumber;
private final int guests;
private final int price;
public Event(int eventNumber, int guests) {
this.eventNumber = eventNumber;
this.guests = guests;
this.price = calculatePrice();
}
public Event() {
this(0, 0);
}
private int calculatePrice() {
return PRICE_PER_GUEST * guests;
}
public int getPrice() {
return price;
}
public int getEventNumber() {
return eventNumber;
}
public int getGuests() {
return guests;
}
// EventDemo.java
package event;
import java.util.Scanner;
public class EventDemo {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
// Create an Event object with default values
Event defaultEvent = new Event();
System.out.println("Default event details:");
System.out.println("Event number: " + defaultEvent.getEventNumber());
System.out.println("Number of guests: " + defaultEvent.getGuests());
System.out.println("Price per guest: $" + defaultEvent.getPrice());
// Create an Event object with user input
System.out.println("Enter event number:");
int eventNumber = input.nextInt();
System.out.println("Enter number of guests:");
int guests = input.nextInt();
Event userEvent = new Event(eventNumber, guests);
System.out.println("Event details:");
System.out.println("Event number: " + userEvent.getEventNumber());
System.out.println("Number of guests: " + userEvent.getGuests());
System.out.println("Price per guest: $" + userEvent.getPrice());
}
The `Event` class now contains two overloaded constructors, one that accepts an event number and number of guests as parameters and another default constructor that passes "A000" and 0 to the two-parameter constructor.
The `EventDemo` class demonstrates how to create two `Event` objects, one with default values and another with user input. It then displays the details of both objects.
Use a programmable calculator or computer (or the sum command on a CAS) to estimate 1 + xe−y R dA where [0, 1] × [0, 1]. Use the Midpoint Rule with the following numbers of squares of equal size: 1, 4, 16, 64, 256, and 1024. (Round your answers to six decimal places.)
Answer:
1.141606
1.143191
1.143535
1.143617
1.143637
1.143642
Explanation:
Suppose that a program performs an intermixed sequence of (stack) push and pop operations. The push operations put the integers 0 through 9 in order onto the stack; the pop operations print out the return values. Which of the following sequence(s) could not occur?
a. 4321098765
b. 2143658790
c. 0465381729
d. 4687532901
Answer:
c and d.
c. 0465381729
d. 4687532901
Explanation:
Once an item has been stacked on top of another item, there
is no way to pop them in a different order.
write a program that reads an integer and displays, using asterisks a filled and hollow square, placed next to each other. for example if side length is 5 the program should display like so.
This program prints a filled and hollow square.
Enter the length of a side: 5
***** *****
***** * *
***** * *
***** * *
***** *****
A Python program can be written to read an integer that is then used to print out two squares of that side length with asterisks, one filled and one hollow. The provided Python code uses nested loops and conditionals to generate the squares accurately.
Explanation:To complete your request, we would need to write a program to read an integer input and utilize this integer value to generate two squares with asterisks, one filled and one hollow. Here is a simple Python program:
def print_squares(n):This program first prints a filled square and a hollow square using conditionals to distinguish between the edge and inner positions of the squares.
Learn more about Python programming here:https://brainly.com/question/33469770
#SPJ3
What is ISP? What is peering? Understand that carriers usually don’t charge one another for peering. Instead, "the money is made" in the ISP business by charging the end-points in a network—the customer organizations and end users that an ISP connects to the Internet.
Answer:
ISP: internet service provider the company that is able to provide you with access to the Internet services. Example: AT & T.
Peering: a telecommunication method that allows two networks to connect and exchange traffic directly without having to pay a third party to carry traffic across.
Explanation:
An Internet Service Provider (ISP) facilitates access to the Internet for its customers, varying in services and size. Peering between ISPs allows direct data interchange without costs, contrasting with ISP revenue generated from charging end-users and organizations for Internet access. The concept of net neutrality emphasizes equal treatment of all Internet data to prevent service disparities.
Explanation:An Internet Service Provider (ISP) is a company that provides customers with access to the Internet. ISPs can vary in size from small community providers to large multinational companies, and they offer different types of Internet connections such as broadband, DSL, and fiber optics. A critical aspect of the ISP business is the practice of peering, which is a direct interconnection between the networks of two ISPs, allowing them to exchange traffic without the involvement of a third party. Peering is typically done without the exchange of money between ISPs; instead, they benefit mutually from the direct flow of data between their networks. This practice contrasts with the way ISPs earn revenue, which is primarily through charging end-users and customer organizations for Internet access. The fees for Internet access can vary depending on the speed and reliability of the connection provided. As the Internet's infrastructure continues to evolve, debates such as those surrounding net neutrality have emerged, stressing the importance of treating all Internet traffic equally to avoid creating disparities between users based on their ability to pay for faster services.
ular RSA modulus sizes are 1024, 2048, 3072 and 4092 bits. How many random odd integers do we have to test on average until we expect to find one that is a prime
Answer:
In RSA, the bit size n of the public modulus N is often of the form n=c⋅2k with c a small odd integer. c=1 (n=512, 1024, 2048, 4096.. bit) is most common, but c=3 (n=768, 1536, 3072.. bit) and c=5 (n=1280..) are common. One reason for this is simply to limit the number of possibilities, and similar progressions are found everywhere in cryptography, and often in computers where binary rules (e.g. size of RAM).
The difficulty of factoring (thus, as far as we know, the security of RSA in the absence of side-channel and padding attacks) grows smoothly with n. But the difficulty of computing the RSA public and private functions grows largely stepwise as n increases (more on why in the next paragraph). The values of n just below a step is thus more attractive than the value just above a step: they are about as secure, but the later is more difficult/slow in actual use. And, not coincidentally, the common RSA modulus sizes are just below such steps.
One major factor creating a step is when one more word/storage unit becomes required to store a number. When the storage unit is b-bit, there is such step every b bits for the RSA public function x↦xemodN; and a step every r⋅b bits for the RSA private function x↦xdmodN, with r=1 for the naïve implementation, and r equal to the number of factors of N when using the CRT with factors of equal bit size (most usually r=2, but I have heard of plans up to r=8). On any modern general-purpose CPU suitable for RSA, b is a power of two and at the very least 25, creating a strong incentive that n is at least a multiple of 26 (r=2 is common, and the only reasonable choice for n below about a thousand).
Note: n=1984=31⋅26 is not unseen in the field of Smart Cards, because the next multiple of 26 would break the equivalent of a sound barrier in a common transport protocol, ISO/IEC 7816-3 T=0. For a list of common RSA modulus size in this field, search LENGTH_RSA_1984.
As an aside, it is significantly simpler to code quotient estimation in Euclidian division (something much used in RSA) when the number of bits of the divisor is known in advance. This creates an incentive to reduce the number of possible bit sizes for the modulus. The two simplest cases are when the number of bits is a multiple of b, and one more than a multiple of b; the former won.
UNIX treats file directories in the same fashion as files; that is, both are defined by the same type of data structure, called an inode. As with files, directories include a nine-bit protection string. If care is not taken, this can create access control problems. For example, consider a file with protection mode 644 (octal) contained in a directory with protection mode 730. How might the file be compromised in this case? What are the limitations? Hint: This is a relatively famous exploit. Give specific commands that will accomplish the exploit.
Answer:
The explanation is listed in the Explanation section below.
Explanation:
All the UNIX directories or files have their summary contained in such an' inode ' format. The inode includes data regarding the current location and size of a file, final release duration, last change time, authorization, etc. Directories are always shown as files and also have an inode connected to them.Unless the folder is big, a list of references to existing data points is implicitly pointed to by inode.
It is composed of the relevant areas:
Type of file.Connect number.Position of data throughout the file.Admin privileges to access files.The system or device command is mkdir and rm (mode) to build directory.
Data driven processes: Select one: a. are heavily based on intuition b. rely heavily on the experience of the process owners c. are based on statistical data, measurement and metrics d. do NOT rely on mathematical models
Answer:
c. are based on statistical data, measurement and metrics
Explanation:
Data driven process are process that are not based on intuition but rather are based on data. This data serves as evidence to back a decision that is to be taken. It therefore means that, whatever decision that will be taken, such a decision will be based on the data presented.
Data driven processes: c. are based on statistical data, measurement and metrics .
Data-driven processes emphasize making decisions and formulating strategies based on empirical evidence and statistical data rather than relying on intuition or personal experience. Such processes incorporate rigorous measurement and metrics to objectively evaluate performance and outcomes, effectively minimizing bias.
Therefore, the correct answer is:
c. are based on statistical data, measurement and metrics