Basic
How do I catch multiple exceptions?
If a try block can throw several kind of exceptions, and you want to handle each exception differently, you can put several catch blocks to handle it. package org.kodejava.basic; public class MultipleCatchExample { public static void main(String[] args) { int[] numbers1 = {1, 2, 3, 4, 5}; int[] numbers2 = {1, 2, 3, 4, 5, […]
Artikel oleh Wayan · Sumber: Kode Java ↗
Terjemahan artikel ini belum tersedia. Isi asli ditampilkan.
If a try block can throw several kind of exceptions, and you want to handle each exception differently, you can put several catch blocks to handle it.
package org.kodejava.basic;
public class MultipleCatchExample {
public static void main(String[] args) {
int[] numbers1 = {1, 2, 3, 4, 5};
int[] numbers2 = {1, 2, 3, 4, 5, 6};
try {
// This line throws an ArrayIndexOutOfBoundsException
MultipleCatchExample.printResult(numbers1);
// This line throws an ArithmeticException
MultipleCatchExample.printResult(numbers2);
} catch (ArithmeticException e) {
e.printStackTrace();
} catch (ArrayIndexOutOfBoundsException e) {
e.printStackTrace();
} finally {
System.out.println("Finally block is always executed.");
}
}
/**
* Divide the given first number by the second number.
*
* @param x the first number.
* @param y the second number.
* @return the result of division.
*/
private static int divide(int x, int y) {
return x / y;
}
/**
* Print the output result of divide operation by calling the
* divide() method.
*
* @param numbers integer arrays of the divided number
* @throws ArrayIndexOutOfBoundsException when an exception
* occurs.
*/
private static void printResult(int[] numbers) {
int x, z, y = 1;
for (int i = 0; i < 6; i++) {
x = numbers[i];
if (i == 5) {
y = 0;
}
z = MultipleCatchExample.divide(x, y);
System.out.println("z = " + z);
}
}
}
