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Paladinen [302]
2 years ago
5

The program DebugTwo2.cs has syntax and/or logical errors. Determine the problem(s) and fix the program.// This program greets t

he user// and multiplies two entered valuesusing System;using static System.Console;class DebugTwo2{static void main(){string name;string firstString, secondSting;int first, second, product;Write("Enter your name >> );name = ReadLine;Write("Hello, {0}! Enter an integer >> ", name);firstString = ReadLine();first = ConvertToInt32(firstString);Write("Enter another integer >> ");secondString = Readline();second = Convert.ToInt(secondString);product = first * second;WriteLine("Thank you, {1}. The product of {2} and {3} is {4}", name, first, second, product);}}
Computers and Technology
1 answer:
Elenna [48]2 years ago
5 0

Answer:

The corrected code is as follows:

using System;

using static System.Console;

class DebugTwo2{

     static void Main(string[] args){

       string name;string firstString, secondString;

       int first, second, product;

       Write("Enter your name >> ");

       name = ReadLine();

       Write("Hello, {0}! Enter an integer >> ", name);

       firstString = ReadLine();

       first = Convert.ToInt32(firstString);

       Write("Enter another integer >> ");

       secondString = ReadLine();

       second = Convert.ToInt32(secondString);

       product = first * second;

       Write("Thank you, {0}. The product of {1} and {2} is {3}", name,first, second, product);

   }

}

Explanation:

       string name;string firstString, secondString; ----- Replace secondSting with secondString,

<em>        int first, second, product;</em>

       Write("Enter your name >> "); --- Here, the quotes must be closed with corresponding quotes "

       name = ReadLine(); The syntax of readLine is incorrect without the () brackets

<em>        Write("Hello, {0}! Enter an integer >> ", name);</em>

<em>        firstString = ReadLine();</em>

       first = Convert.ToInt32(firstString); There is a dot between Convert and ToInt32

<em>        Write("Enter another integer >> ");</em>

       secondString = ReadLine(); --- Readline() should be written as ReadLine()

<em>        second = Convert.ToInt32(secondString);</em>

<em>        product = first * second;</em>

       Write("Thank you, {0}. The product of {1} and {2} is {3}", name,first, second, product); --- The formats in {} should start from 0 because 0 has already been initialized for variable name

   }

}

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given:an int variable k,an int array currentMembers that has been declared and initialized,an int variable memberID that has bee
Oksi-84 [34.3K]

Answer:

// The code segment is written in C++ programming language

// The code segment goes as follows

for (k = 0; k < nMembers; k++)

{

//check if memberID can be found in currentMembers

if (currentMembers[k] == memberID){

// If yes,

// assigns true to isAMember

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}

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}

// End of segment:

The following assumption were made in the code segment above.

There exists

1. An already declared and initialised int array currentMembers.

2. An already initialised int variable memberID

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7 0
2 years ago
Write a for loop that prints the odd integers 11 through 121 inclusive, separated by spaces.
mixas84 [53]
The code for the above problem is:
<span>/* package whatever; // don't place package name! */ 

import java.util.*;
import java.lang.*;
import java.io.*; 

/* Name of the class has to be "Main" only if the class is public. */

class Ideone{ public static void main (String[] args) throws java.lang.Exception 
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4 0
2 years ago
Create the class named SortFile. The class will have a single attribute, which is an ArrayList of integers. The class will have
S_A_V [24]

Answer:

See explaination

Explanation:

//SortFile.java

import java.io.File;

import java.io.FileNotFoundException;

import java.util.ArrayList;

import java.util.Arrays;

import java.util.Scanner;

public class SortFile {

// array list of integers to store the numbers

private ArrayList<Integer> list;

// default constructor

public SortFile() {

// initializing list

list = new ArrayList<Integer>();

// reading file name using a scanner

Scanner sc = new Scanner(System.in);

System.out.print("Enter file name: ");

String filename = sc.nextLine();

// opening file inside a try catch block. if you want to throw exception

// instead, just remove try-catch block and add throws

// FileNotFoundException to the method signature.

try {

// opening file

sc = new Scanner(new File(filename));

// looping and reading each integer from file to list

while (sc.hasNextInt()) {

list.add(sc.nextInt());

}

// closing file

sc.close();

} catch (FileNotFoundException e) {

// exception occurred.

System.err.println(e);

}

}

// constructor taking a file name

public SortFile(String filename) {

// initializing list

list = new ArrayList<Integer>();

// opening file inside a try catch block. if you want to throw exception

// instead, just remove try-catch block and add throws

// FileNotFoundException to the method signature.

try {

Scanner sc = new Scanner(new File(filename));

// looping and reading each integer from file to list

while (sc.hasNextInt()) {

list.add(sc.nextInt());

}

sc.close();

} catch (FileNotFoundException e) {

System.err.println(e);

}

}

// method to return a sorted array of integers containing unique elements

// from list

public int[] sort() {

// creating an array of list.size() size

int arr[] = new int[list.size()];

// count of unique (non duplicate) numbers

int count = 0;

// looping through the list

for (int i = 0; i < list.size(); i++) {

// fetching element

int element = list.get(i);

// if this element does not appear before on list, adding to arr at

// index=count and incrementing count

if (!list.subList(0, i).contains(element)) {

arr[count++] = element;

}

}

// now reducing the size of arr to count, so that there are no blank

// locations

arr = Arrays.copyOf(arr, count);

// using selection sort algorithm to sort the array

int index = 0;

// loops until index is equal to the array length. i.e the whole array

// is sorted

while (index < arr.length) {

// finding the index of biggest element in unsorted array

// initially assuming index as the index of biggest element

int index_max = index;

// looping through the unsorted part of array

for (int i = index + 1; i < arr.length; i++) {

// if current element is bigger than element at index_max,

// updating index_max

if (arr[i] > arr[index_max]) {

index_max = i;

}

}

// now we simply swap elements at index_max and index

int temp = arr[index_max];

arr[index_max] = arr[index];

arr[index] = temp;

// updating index

index++;

// now elements from 0 to index-1 are sorted. this will continue

// until whole array is sorted

}

//returning sorted array

return arr;

}

//code for test-run

public static void main(String[] args) {

//initializing SortFile using default constructor

SortFile sortFile = new SortFile();

//displaying sorted array

System.out.println(Arrays.toString(sortFile.sort()));

}

6 0
2 years ago
Given 4 integers, output their product and their average, using integer arithmetic.
solmaris [256]

Answer:

see explaination

Explanation:

Part 1:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

int num3;

int num4;

int avg=0, pro=1;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

num3 = scnr.nextInt();

num4 = scnr.nextInt();

avg = (num1+num2+num3+num4)/4;

pro = num1*num2*num3*num4;

System.out.println(pro+" "+avg);

}

}

------------------------------------------------------------------

Part 2:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

int num3;

int num4;

double avg=0, pro=1; //using double to store floating point numbers.

num1 = scnr.nextInt();

num2 = scnr.nextInt();

num3 = scnr.nextInt();

num4 = scnr.nextInt();

avg = (num1+num2+num3+num4)/4.0; //if avg is declared as a float, then use 4.0f

pro = num1*num2*num3*num4;

System.out.println((int)pro+" "+(int)avg); //using type conversion only integer part

System.out.printf("%.3f %.3f\n",pro,avg);// \n is for newline

}

}

8 0
2 years ago
A small company is moving towards sharing printers to reduce the number of printers used within the company. The technician has
Misha Larkins [42]

Answer:

Security risk associated with sharing a printer are

  1. Printer Attacks
  2. Theft
  3. Breach of data
  4. Vulnerable Network

Explanation:

Printer Attacks

A network printer can be used for a DDoS attack.As printer are not very secured and are a weak link in network these can be easily exploited by the hackers for any kind of malicious activities and even lanching a DDoS attack.

DDoS attack is denial of service attack in which network is flooded with malicious traffic which cause it to choke and make it inaccessible for users.

Theft

Physical theft of document can be an issue.Anyone can just took printed pages from printer tray by any one.

Breach of Data

The documents which are printed are usually stored in printer cache for some time which can be accessed by any one connected to the network. Any document containing confidential information which are printed on network printer can fall in wrong hands.

Vulnerable Network

As mentioned a single unsecured network printer can pose great threat to entire network as it can provide a way into the network.

8 0
2 years ago
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