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Home/ Questions/Q 7529191
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T04:35:37+00:00 2026-05-30T04:35:37+00:00

My sample code is as follows: public class ExceptionsDemo { public static void main(String[]

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My sample code is as follows:

public class ExceptionsDemo {

    public static void main(String[] args) {
        try {
            int arr[]={1,2,3,4,5,6,7,8,9,10};
            for(int i=arr.length;i<10;i++){
                if(i%2==0){
                    System.out.println("i =" + i);
                    throw new Exception();
                }            
            }
        } catch (Exception e) {
            System.err.println("An exception was thrown");
        }            
    }
}

My requirement is that, after the exception is caught I want to process the remaining elements of the array. How can I do this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T04:35:38+00:00Added an answer on May 30, 2026 at 4:35 am

    Your code should look as follows:

    public class ExceptionsDemo {
    
        public static void main(String[] args) {
            for (int i=args.length;i<10;i++){
                try {
                    if(i%2==0){
                        System.out.println("i =" + i);
                        throw new Exception();  // stuff that might throw
                    }
                } catch (Exception e) {
                    System.err.println("An exception was thrown");
                }
            }
        }
    }
    
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