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(Solved): 2.7 LAB: Exception handling to detect input String vs. Integer The given program reads a list of sin ...



2.7 LAB: Exception handling to detect input String vs. Integer The given program reads a list of single-word first names and ages (ending with -1), and outputs that list with the age incremented. The program fails and throws an exception if the second input on a line is a String rather than an Integer. At FIXME in the code, add a try/catch statement to catch java.util.InputMismatchException, and output 0 for the age. Ex: If the input is: Lee 18 Lua 21 Mary Beth 19 Stu 33 -1 then the output is: Lee 19 Lua 22 Mary 0 Stu 34 LAB ACTIVITY 2.7.1: LAB: Exception handling to detect input String vs. Integer

Java Code:

import java.util.Scanner;
import java.util.InputMismatchException;

public class NameAgeChecker {
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);

String inputName;
int age;

inputName = scnr.next();
while (!inputName.equals("-1")) {
// FIXME: The following line will throw an InputMismatchException.
// Insert a try/catch statement to catch the exception.
try{
age = scnr.nextInt();
System.out.println(inputName + " " + (age + 1));
}
catch (InputMismatchException e) {
scnr.nextLine();
}

inputName = scnr.next();
}
}
}



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