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Start your free trialNeelesh Tewani
1,239 Pointsi 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"); } } }
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
15,353 PointsHey 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
Treehouse TeacherKen Alger
Treehouse TeacherAnother great example as to why single letter variable names can be confusing. I tend to call my exceptions
exception
.