Java Multiple Exceptions
Multiple Exceptions
Sometimes, different errors (exceptions) can happen in the same try block.
You can handle them with multiple catch blocks.
One try, Many catch
You can add more than one catch block, and Java will run the first one that matches the thrown exception type:
Example
public class Main {
public static void main(String[] args) {
try {
int[] numbers = {1, 2, 3};
System.out.println(numbers[10]); // ArrayIndexOutOfBoundsException
int result = 10 / 0; // ArithmeticException
}
catch (ArrayIndexOutOfBoundsException e) {
System.out.println("Array index does not exist.");
}
catch (ArithmeticException e) {
System.out.println("Cannot divide by zero.");
}
catch (Exception e) {
System.out.println("Something else went wrong.");
}
}
}
Result:
Array index does not exist.
Explanation: Only the first exception (ArrayIndexOutOfBoundsException) is thrown, so only the first matching catch runs.
Order Matters
You should always put more specific exceptions first, and general ones later. Otherwise, the general catch will grab the error and the specific ones will never run:
Example (bad order)
try {
int result = 10 / 0;
}
catch (Exception e) {
System.out.println("General error.");
}
catch (ArithmeticException e) {
// This will never be reached
System.out.println("Divide by zero.");
}
Tip: Always put Exception (the general one)
at the end.
Multi-Catch
Since Java 7, you can catch multiple exceptions in one catch block using the | symbol.
This is useful when different exceptions should be handled in the same way, so you don't have to repeat code:
Example
try {
int result = 10 / 0;
int[] numbers = {1, 2, 3};
System.out.println(numbers[10]);
}
catch (ArithmeticException | ArrayIndexOutOfBoundsException e) {
System.out.println("Math error or array error occurred.");
}
