尝试打印多个 catch 语句

Question

In the following code, I am trying to print multiple catch statements but I am getting only one. As far as I understand that the order is prioritized ie the first catch statement matching will be printed. But I want to print both relevant statements. is there any way for this?

    class Example2{
    public static void main(String args[]){
     try{
         int a[]=new int[7];
         a[10]=30/0;
         System.out.println("First print statement in try block");
     }
     catch(ArithmeticException e){
        System.out.println("Warning: ArithmeticException");
     }
     catch(ArrayIndexOutOfBoundsException e){
        System.out.println("Warning: ArrayIndexOutOfBoundsException");
     }
     catch(Exception e){
        System.out.println("Warning: Some Other exception");
     }
   System.out.println("Out of try-catch block...");
  }
}

I want both, out of bound and arithmetic statements to be printed. Is there any way?

Answer 1

The issue here isn't prioritization of the catch blocks. First, you attempt to divide 30/0 , and generate an ArithmeticException . The ArrayIndexOutOfBounds exception will never be generated because there's never a value for you to try to assign to a[10] .

Answer 2

异常只匹配一个 catch 块。

Answer 3

You need to merge those catch statements, because only one is fired at a

class Example2{
    public void main(String args[]){
        try{
            int a[]=new int[7];
            a[10]=30/0;
            System.out.println("First print statement in try block");
        } catch(ArithmeticException | ArrayIndexOutOfBoundsException  e) {

        }
        System.out.println("Out of try-catch block...");
    }
}

And in the catck block, you can then work with the exception.

Answer 4

There is one way to print both the exceptions by using nested try statements as shown below. Otherwise, it should not be necessary to print all the exceptions.

class ExceptionHandling{
    public static void main(String[] args){
        try{
            try{
                String s=null;
                System.out.println(s.length());
            }
            
            catch(NullPointerException e){
                System.out.println(e);
            }   
            
            int a=4/0;
        }
        catch(ArithmeticException e){
            System.out.println(e);
        }
    }
}