I want you to make this Java code bulletproof, by making it output…
Question Answered step-by-step I want you to make this Java code bulletproof, by making it output… I want you to make this Java code bulletproof, by making it output an error message if the user inputs a letter or symbol, preferably use toString, parseInt, and a try catch. Thank you. /*This program reads a positive integer and then checks to see whether or notthe integer is divisible by 11. For that divisibility test, use the followingalgorithm, based on one given by the mathematician Lewis Carroll.*/public class Divisible_name{ public static void main (String[] args) { //Declares variables int numberIn, numberB, numberC, numberD, numberDup; numberIn = 0; //While loop which works when user input is not 99999 while (numberIn != 99999) { System.out.println (“nEnter a positive number or enter ‘99999’ to end the program: “); numberIn = In.getInt (); numberDup = numberIn; //While loop which implements the formula, and determines the output while (numberIn > 9 && numberIn != 0 && numberIn > 0 && numberIn != 99999) { numberB = numberIn / 10; numberC = numberIn % 10; numberIn = numberB – numberC; System.out.println (numberIn); } //If input is 99999, program end message is printed if (numberIn == 99999) System.out.println (“The program has ended.”); //If input is 0 number is divisible by 11 if (numberIn == 0) { System.out.println (“The number is divisible by 11.”); } //Outputs a space, so that the else does not include the sentinel else if (numberIn == 99999) { System.out.println (“r”); } //If none of the other cases occur, the number is not divisible else { System.out.println (“The number is NOT divisible by 11.”); } } }} Computer Science Engineering & Technology Java Programming COM SCI ICS3U1 Share QuestionEmailCopy link Comments (0)
