Java Java Objects Harnessing the Power of Objects Handling Exceptions

can any one help me to explain that quiz because i din't understood it.

help me please

Example.java
class Example {

  public static void main(String[] args) {
    GoKart kart = new GoKart("purple");
    if (kart.isBatteryEmpty()) {
      System.out.println("The battery is empty");
    }
    kart.drive(42); 
  }
  if (kart>barcount) {
  throw new IllegalArgumentException ("never");
}
}
GoKart.java
class GoKart {
  public static final int MAX_BARS = 8;
  private int barCount;
  private String color;
  private int lapsDriven;

  public GoKart(String color) {
    this.color = color;
  }

  public String getColor() {
    return color;
  }

  public void charge() {
    barCount = MAX_BARS;
  }

  public boolean isBatteryEmpty() {
    return barCount == 0;
  }

  public boolean isFullyCharged() {
    return MAX_BARS == barCount;
  }

  public void drive() {
    drive(1);
  }

  public void drive(int laps) {
    if (laps > barCount) {
      throw new IllegalArgumentException("Not enough battery remains");
    }
    lapsDriven += laps;
    barCount -= laps;
  }

}
James Han
James Han
9,360 Points

I believe it is asking you to "try" driving the kart for 42 laps and if the laps exceeds MAX_BAR available (each lap is 1 bar), then catch the IllegalArgumentException thrown.

public static void main(String[] args) {
    GoKart kart = new GoKart("purple");
    if (kart.isBatteryEmpty()) {
      System.out.println("The battery is empty");
    }
    try {
      kart.drive(42); 
    } catch (IllegaleArgumentException iae) {
      System.out.printf("%s", iae.getMessage());
    }
  }

I believe something like that will work.

1 Answer

Balazs Pukli
Balazs Pukli
46,041 Points

When it comes to exceptions, you throw them in one place, and catch them in another. Where you call the function that may or may not throw one, you need to put that funciton call into a try-catch block.

    try{
    kart.drive(42);
    }
    catch( IllegalArgumentException exception ){
        System.out.println(exception.getMessage());
    }