Why do I get compile error? using try and catch

Question

public static void main(String[] args) {
        // TODO Auto-generated method stub

        int x = 0, y = 10;
        try {
            y /=x;
        }
        System.out.print(" / by 0");
        catch(exception e) {
            System.out.print("error");
        }
}

Using try and catch, I have built code above and the output is compile error. The output I expected was "error" since a number divided by zero gives an ArithmeticException. Why do I get compile error above?

Answer 1

You can't seperate the try from the catch block. This got you the compilation error. Correct code would be:

public static void main(String[] args) {
    int x = 0, y = 10;
    try {
        y /= x;
    } catch (Exception e) {
        System.out.print(" / by 0");
        System.out.print("error");
    }
}

Also, you had Exception lowercase, which would have caused another problem.

Answer 2

A catch block is supposed to be immediately followed by try block. Put the print line statement inside catch block. Also, Exception is a class of Java, exception is not.

So, put the catch keyword immediately after the try block (}) ends, or in the next line.