Question

7.6 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

250306.1245598

lab activity

7.6.1: LAB: Exception handling to detect input String vs. Integer

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.
         age = scnr.nextInt();
         System.out.println(inputName + " " + (age + 1));
       
         inputName = scnr.next();
      }
   }
}

Answer 1

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"))) {
            try {
                age = scnr.nextInt();
            } catch (InputMismatchException e) {
                age = -1;
                scnr.nextLine();
            }
            System.out.println(inputName + " " + (age + 1));
            inputName = scnr.next();
        }
    }
}