Question about how to exit program with exception

Assume that I want to exit a console program if the user entered the char f, and in any time of the program. The user is supposed to enter some info but I want for each step he entering the input to be able to stop开发者_如何学C all the operation if he entered "f"? How can I do that? Should it be something like:

try
{
 if (userchoice.equals("F"))
  {
    throw new exception e;
  }
}
catch (exception e)
{
  System.exit(1);
}  

Thanks

---

You can throw the exception, unless it is caught it will cause the current thread to die.

if ("f".equalsCaseIgnore(userchoice))
    throw new IllegalArgumentException("Option "+userchoice+" not allowed.");

---

Here's the correct syntax:

try {
  if (userchoice.equals("F")) {
     throw new Exception();
  }
} catch (Exception e){
  System.exit(1);
}

---

Hint:

• Scanner class
• System.exit();

---

The input of the char "f" is expected behavior and so throwing an exception may be wrong way.

Encapsulate your input in a method which is responsible to handle the input and decided behavior.

Just call system.exit() here if the user entered "f" or call an exit method that does the work.

---

Do the following.

• Read the input from the command line using classes like BufferedReader or Scanner.
• Check for the character "f" from the i/p'ed string.
• Throw your exception using throw new MyException();
• Catch the exception in the catch block and terminate it with System.exit(1);

If you require "f" or "F", use equalsIgnoreCase() function.

Try something like this,

  try{
    if (userchoice.equals("F")) {
         throw new MyException;
       }
    }
 catch (MyException e) {
  System.out.println("MyException caught because i/p character was F" + e);
  System.exit(1);
  }