JuniorTheoryCommon

What happens when this program is compiled and run?

JAVA

public class Main {
    public static void main(String[] args) {
        int d = 0;
        try {
            System.out.println(10 / d);
        } catch (Exception e) {
            System.out.println("error");
        } catch (ArithmeticException e) {
            System.out.println("zero");
        }
    }
}

It does not compile

Expecting error: Java does take the first matching catch at run time, but here the compiler sees that the second one can never run and refuses the program.
Expecting zero: Java does not pick the most specific catch; it checks them top to bottom, so a subclass must come before its parent.

How do you fix the program so that it prints zero?
What does catch (ArithmeticException | NumberFormatException e) do?

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JuniorTheoryCommon

What happens when this program is compiled and run?

JAVA

public class Main {
    public static void main(String[] args) {
        int d = 0;
        try {
            System.out.println(10 / d);
        } catch (Exception e) {
            System.out.println("error");
        } catch (ArithmeticException e) {
            System.out.println("zero");
        }
    }
}

It does not compile

Expecting error: Java does take the first matching catch at run time, but here the compiler sees that the second one can never run and refuses the program.
Expecting zero: Java does not pick the most specific catch; it checks them top to bottom, so a subclass must come before its parent.

How do you fix the program so that it prints zero?
What does catch (ArithmeticException | NumberFormatException e) do?

Back to questions Open topic in roadmap →Answer it yourself — Quiz