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GarryVolchara [31]
3 years ago
7

Assume that processor refers to an object that provides a void method named process that takes no arguments. As it happens, the

process method may throw one of several exceptions. Write some code that invokes the process method provided by the object associated with processor and arrange matters so that your code causes any exception thrown by process to be ignored. Hint: use the catch (Exception ex) and do nothing under the catch clause.
Computers and Technology
1 answer:
Aloiza [94]3 years ago
5 0

Answer:

Following are the code to the given question:

try//defining a try block

{

processor.process();//defining an object processor that calls process method

}

catch(Exception e)//defining a catch block

{

}

Explanation:

In this question, the 'Try' and 'catch' block is used in which both the keywords are used to represent exceptions managed during runtime due to information or code errors. This try box was its code block which includes errors. A message queue catches the block errors and examines these.

In the try block, a method "process" is used which is create the object processor that calls the method.

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Write the definition of a method named countPos that receives a reference to a Scanner object associated with a stream of input
NNADVOKAT [17]

Answer:

Here is the method countPos    

  static int countPos(Scanner input){ //method that takes a reference to a Scanner object

   int counter=0;  // to count the positive integers  

   int number; //to store each integer value

   if(input.hasNextInt()){ //checks if next token in this scanner input is an integer value

       number=input.nextInt(); //nextInt() method of a Scanner object reads in a string of digits and converts them into an int type and stores it into number variable

       counter=countPos(input); //calls method by passing the Scanner object and stores it in counter variable

       if(number>0) //if the value is a positive number

           counter++;    } //adds 1 to the counter variable each time a positive input value is enountered

   else    { //if value is not a positive integer

       return counter;    } //returns the value of counter

   return counter; } //returns the total count of the positive numbers

Explanation:

Here is the complete program:

import java.util.Scanner; //to take input from user

public class Main { //class name

//comments with each line of method below are given in Answer section

static int countPos(Scanner input){

   int counter=0;

   int number;

   if(input.hasNextInt()){

       number=input.nextInt();

       counter=countPos(input);

       if(number>0)

           counter++;    }

   else    {

       return counter;    }

   return counter; }

public static void main(String[] args) { //start of main function

       System.out.println("Number of positive integers: " + countPos(new Scanner(System.in)));  } } //prints the number of positive integers by calling countPos method and passing Scanner object to it

The program uses hasNextInt method that returns the next token (next input value) and if condition checks using this method if the input value is an integer. nextInt() keeps scanning the next token of the input as an integer. If the input number is a positive number then the counter variable is incremented to 1 otherwise not. If the use enters anything other than an integer value then the program stops and returns the total number of positive integers input by the user. This technique is used in order to avoid using any loop. The program and its output is attached

3 0
3 years ago
The acronym is used to define the process that allows multiple devices to share a single routable ip address.
blondinia [14]
The answer is Mac Address
media access control address<span> </span>
4 0
3 years ago
A school operated for the express purpose of giving its students the skills
miss Akunina [59]

Answer:

vocational school

Explanation:

These vocational schools are institution postsecondary and job training, these schools offer quickly programs, but with the knowledge to work in a company, if we're talking about technology programs we can consider, programming, network administrator, technician.

These institutes are for people like:

  • People without work experience
  • People want to start a new career
  • People want to reenter to the working market
3 0
3 years ago
Complete function PrintPopcornTime(), with int parameter bagOunces, and void return type. If bagOunces is less than 3, print "To
SVETLANKA909090 [29]

Answer:

<h2>Function 1:</h2>

#include <stdio.h> //for using input output functions

// start of the function PrintPopcornTime body having integer variable //bagOunces as parameter

void PrintPopcornTime(int bagOunces){

if (bagOunces < 3){ //if value of bagOunces is less than 3

 printf("Too small"); //displays Too small message in output

 printf("\n"); } //prints a new line

//the following else if part will execute when the above IF condition evaluates to //false and the value of bagOunces is greater than 10

else if (bagOunces > 10){

    printf("Too large"); //displays the message:  Too large in output

    printf("\n"); //prints a new line }

/*the following else  part will execute when the above If and else if conditions evaluate to false and the value of bagOunces is neither less than 3 nor greater than 10 */

else {

/* The following three commented statements can be used to store the value of bagOunces * 6 into result variable and then print statement to print the value of result. The other option is to use one print statement printf("%d",bagOunces * 6) instead */

    //int result;

    //result = bagOunces * 6;

    //printf("%d",result);

 printf("%d",bagOunces * 6);  /multiplies value of bagOunces  to 6

 printf(" seconds");

// seconds is followed with the value of bagOunces * 6

 printf("\n"); }} //prints a new line

int main(){ //start of main() function body

int userOunces; //declares integer variable userOunces

scanf("%d", &userOunces); //reads input value of userOunces

PrintPopcornTime(userOunces);

//calls PrintPopcornTime function passing the value in userOunces

return 0; }

Explanation:

<h2>Function 2:  </h2>

#include <stdio.h> //header file to use input output functions

// start of the function PrintShampooInstructions body having integer variable numCycles as parameter

void PrintShampooInstructions(int numCycles){

if(numCycles < 1){

//if conditions checks value of numCycles is less than 1 or not

printf("Too few."); //prints Too few in output if the above condition is true

printf("\n"); } //prints a new line

//else if part is executed when the if condition is false and else if  checks //value of numCycles is greater than 4 or not

else if(numCycles > 4){

//prints Too many in output if the above condition is true

printf("Too many.");

printf("\n"); } //prints a new line

//else part is executed when the if and else if conditions are false

else{

//prints "N: Lather and rinse." numCycles times, where N is the cycle //number, followed by Done

for(int N = 1; N <= numCycles; N++){

printf("%d",N);

printf(": Lather and rinse. \n");}

printf("Done.");

printf("\n");} }

int main() //start of the main() function body

{    int userCycles; //declares integer variable userCycles

   scanf("%d", &userCycles); //reads the input value into userCycles

   PrintShampooInstructions(userCycles);

//calls PrintShampooInstructions function passing the value in userCycles

   return 0;}

I will explain the for loop used in PrintShampooInstructions() function. The loop has a variableN  which is initialized to 1. The loop checks if the value of N is less than or equal to the value of numCycles. Lets say the value of numCycles = 2. So the condition evaluates to true as N<numCycles  which means 1<2. So the program control enters the body of loop. The loop body has following statements. printf("%d",N); prints the value of N followed by

printf(": Lather and rinse. \n"); which is followed by printf("Done.");

So at first iteration:

printf("%d",N); prints 1 as the value of N is 1

printf(": Lather and rinse. \n");  prints : Lather and rinse and prints a new line \n.

As a whole this line is printed on the screen:

1: Lather and rinse.

Then the value of N is incremented by 1. So N becomes 2 i.e. N = 2.

Now at second iteration:

The loop checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to true as N<numCycles  which means 2=2. So the program control enters the body of loop.

printf("Done."); prints Done after the above two lines.

printf("%d",N); prints 2 as the value of N is 2

printf(": Lather and rinse. \n");  prints : Lather and rinse and prints a new line \n.

As a whole this line is printed on the screen:

2: Lather and rinse.

Then the value of N is incremented by 1. So N becomes 2 i.e. N = 3.

The loop again checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to false as N<numCycles  which means 3>2. So the loop breaks.

Now the next statement is:

printf("Done."); which prints Done on the screen.

So as a whole the following output is displayed on the screen:

1: Lather and rinse.

2: Lather and rinse.

Done.

The programs along with their outputs are attached.

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3 years ago
The action in which a router divides and forwards incoming or outbound message traffic to multiple links is known as
kipiarov [429]
Load balancing is my best guess.
5 0
3 years ago
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