How to make a general exception in Java?

Question

Consider this simple program. The program has two files:

Vehicle.java: class Vehicle { private int speed = 0; private int maxSpeed = 100; public int getSpeed() { return speed; } public int getMaxSpeed() { return maxSpeed; } public void speedUp(int increment) { if(speed + increment > maxSpeed){ // throw exception }else{ speed += increment; } } public void speedDown(int decrement) { if(speed - decrement < 0){ // throw exception }else{ speed -= decrement; } } } 

And HelloWorld.java:

 public class HelloWorld { /** * @param args */ public static void main(String[] args) { Vehicle v1 = new Vehicle(); Vehicle v2 = new Vehicle(); // do something // print something useful, TODO System.out.println(v1.getSpeed()); } } 

As you can see in the first class, I added a comment ("// throw exception"), where I would like to make an exception. Should I define my own class for exceptions, or is there any general Java exception class that I can use?

Answer 1

You can create your own Exception class:

 public class InvalidSpeedException extends Exception { public InvalidSpeedException(String message){ super(message); } } 

In your code:

 throw new InvalidSpeedException("TOO HIGH"); 

Answer 2

You can use IllegalArgumentException:

 public void speedDown(int decrement) { if(speed - decrement < 0){ throw new IllegalArgumentException("Final speed can not be less than zero"); }else{ speed -= decrement; } } 

Answer 3

Well, there are many throw exceptions, but here's how you make an exception:

 throw new IllegalArgumentException("INVALID"); 

Also yes, you can create your own custom exceptions.

Edit: Also note the exceptions. When you throw an exception (like the one above), and you find the exception: the String that you provide in the exception can be accessed by the getMessage() method.

 try{ methodThatThrowsException(); }catch(IllegalArgumentException e) { e.getMessage(); } 

Answer 4

It really depends on what you want to do with this exception after you catch it. If you need to distinguish your exception, you need to create your own Exception . Otherwise, you could simply throw new Exception("message goes here");

Answer 5

The easiest way to do this:

 throw new java.lang.Exception(); 

However, the following lines would not be achievable in your code. So, we have two ways:

---

• throw a general exception at the bottom of the method. \
• create your own exception if you don't want to do 1.

Answer 6

Java has a large number of built-in exceptions for different scenarios.

In this case, you should throw an IllegalArgumentException , since the problem is that the caller passed a bad parameter.

Answer 7

You can define your own exception class extending java.lang.Exception (which should be caught for a checked exception) or extending java.lang.RuntimeException - these exceptions should not be caught. Another solution is to review the Java API and find an appropriate exception that describes your situation: in this particular case, I believe that an IllegalArgumentException would be the best.

Answer 8

It depends, you can throw a more general exception or a more specific exception. For simpler methods, general exceptions are sufficient. If the method is complex, then throwing a more specific exception will be reliable.