为什么 try-catch 没有给出所需的 output？

Question

The following code gives the desired output which is "Arithmetic Exception occurs"

public static void main(String[] args) {

    try {
        int []a = new int[5];
        a[5] = 30/0; 
    } catch(ArithmeticException e) {
        System.out.println("Arithmetic Exception occurs");
    } catch(ArrayIndexOutOfBoundsException e) {
        System.out.println("ArrayIndexOutOfBounds Exception occurs");
    } catch(Exception e) {
        System.out.println("Parent Exception occurs");
    }
    System.out.println("rest of the code");
}

but if I want to get two exceptions at a time, the modified code will be

public static void main(String[] args) {

    try {
        int []a = new int[5];
        a[10] = 30/0; 
    } catch(ArithmeticException e) {
        System.out.println("Arithmetic Exception occurs");
    } catch(ArrayIndexOutOfBoundsException e) {
        System.out.println("ArrayIndexOutOfBounds Exception occurs");
    } catch(Exception e) {
        System.out.println("Parent Exception occurs");
    }
    System.out.println("rest of the code");
}

but instead of giving both, "Arithmetic Exception occurs" & "ArrayIndexOutOfBounds Exception occurs" the output is only "Arithmetic Exception occurs"

Can anyone explain this?

Answer 1

For this line

a[10] = 30/0; 

As soon as 30/0 is evaluated, exception is thrown.

At any point, the code can throw single exception. Even if you catch on the exception and its base class, you can run single catch statement.