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?

Revati S Misra

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