How does JVM "throw" an exception?

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Aryan Kumar · Answer

The Java Virtual Machine (JVM) throws an exception by using the throw keyword. The throw keyword is used to create an exception object and then pass it to the JVM. The JVM then searches the call stack for a matching exception handler. If a matching exception handler is found, the JVM transfers control to the exception handler. The exception handler then handles the exception, which may involve logging the exception, displaying an error message, or terminating the program.

Here is an example of how the throw keyword is used to throw an exception:

public void divide(int x, int y) throws ArithmeticException {
  if (y == 0) {
    throw new ArithmeticException("Division by zero");
  }

  int result = x / y;
  System.out.println(result);
}

In this example, the divide() method throws an ArithmeticException if the y parameter is equal to 0. The throw keyword is used to create an ArithmeticException object and then pass it to the JVM. The JVM then searches the call stack for a matching exception handler. If a matching exception handler is not found, the JVM terminates the program.

The throw keyword can also be used to throw checked exceptions. Checked exceptions are exceptions that must be handled or specified in the method declaration. For example, the IOException exception is a checked exception. If a method throws a checked exception, the method must either handle the exception or specify the exception using the throws keyword.

Here is an example of how the throw keyword is used to throw a checked exception:

public void readFile(String fileName) throws IOException {
  FileInputStream fis = new FileInputStream(fileName);
  int data = fis.read();
  ...
}

In this example, the readFile() method throws an IOException if there is an error reading the file. The throws keyword is used to specify that the readFile() method can throw an IOException exception.

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How does JVM "throw" an exception?

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