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Java Java Objects (Retired) Harnessing the Power of Objects Handling Exceptions

Neelesh Tewani
Neelesh Tewani
1,239 Points

i am not able to do this task

public class Main { public static void main(String[] args) { GoKart kart = new GoKart("yellow"); if (kart.isBatteryEmpty()) { System.out.println("The battery is empty"); } try{ kart.drive(2); } catch(IllegalArgumentException e){ System.out.println("i am warning you"); } } }

Main.java
public class Main {
    public static void main(String[] args) {
        GoKart kart = new GoKart("yellow");
        if (kart.isBatteryEmpty()) {
          System.out.println("The battery is empty");
        }
      try{
        kart.drive(2);
      }
      catch(IllegalArgumentException e){
        System.out.println("i am warning you");
      }
    }
}

1 Answer

Kevin Faust
Kevin Faust
15,353 Points

Hey Neelesh,

I pasted your code and wondered why it wasn't working and that's because the code checker wants you to print out the IllegalArgumentException and nothing else.

System.out.println(e);

That's all it wants you to do. The 'e' is a variable for the illegal argument exception

Hope that cleared your question,

Kevin

Ken Alger
Ken Alger
Treehouse Teacher

Another great example as to why single letter variable names can be confusing. I tend to call my exceptions exception. :smile: