Java
Given an array of 10 names, complete the main method that outputs the name specified by the array index entered by the user. Use a try block to output the name and a catch block to catch any ArrayIndexOutOfBoundsException. Output the message from the exception object if an ArrayIndexOutOfBoundsException is caught. Output the first element in the array if the index is negative or the last element if the index is greater than the size of the array.
Hint: Format the exception outputs using the getMessage() method from the exception object. Do not hard code the exception messages.
Ex: If the input of the program is:
5
the program outputs:
Name: Jane
Ex: If the input of the program is:
12
the program outputs:
Exception! Index 12 out of bounds for length 10
The closest name is: Johnny
Ex: If the input of the program is:
-2
the program outputs:
Exception! Index -2 out of bounds for length 10
The closest name is: Ryley
import java.util.Scanner;
public class LabProgram {
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
String[] names = { "Ryley", "Edan", "Reagan", "Henry", "Caius", "Jane", "Guto", "Sonya", "Tyrese", "Johnny" };
int index;
index = scnr.nextInt();
try {
}
}
}

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MrRoyal MrRoyal · Answer 1 · 2021-10-18T16:06:31+02:00

The program illustrates the use of catching exceptions using try and catch.

The required try and catch block is as follows:

try {

System.out.print(names[index]);

}

catch (ArrayIndexOutOfBoundsException e) {

System.out.println(e.getMessage());

}

The flow of the above code segment is as follows:

First, we begin with the try block

try {

Then the program prints the element at the index

System.out.print(names[index]); }

If the index does not exist in the array (e.g. -1 or 11), then the catch block is executed

catch (ArrayIndexOutOfBoundsException e) {

This prints the appropriate exception

System.out.println(e.getMessage()); }

See attachment for sample run

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