Where is this exception caught and handled?

Question

Aryan Kumar · Answer

The exception in the code snippet below is caught and handled in the catch block that is declared for ArithmeticException.

public class ExceptionHandling {

  public static void main(String[] args) {
    try {
      int x = 10;
      int y = 0;
      int z = x / y;
    } catch (ArithmeticException e) {
      System.out.println("Division by zero");
    }
  }
}

The try block contains the code that could potentially throw an exception. In this case, the code that could throw an exception is the int z = x / y; statement. If the y variable is equal to 0, then the ArithmeticException exception will be thrown.

The catch block is declared for the ArithmeticException exception. This means that if the ArithmeticException exception is thrown, the catch block will be executed. The catch block will then print the message "Division by zero".

If the y variable is not equal to 0, then the ArithmeticException exception will not be thrown. In this case, the try block will execute normally and the catch block will not be executed.

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Where is this exception caught and handled?

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