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Start your free trialRexford Aboagye
3,080 PointsnormalizeDiscount
what am i doing wrong ?
public class Order {
private String itemName;
private int priceInCents;
private String discountCode;
private String normalizeDiscountCode(String discountCode){
for (char res : discountCode.toCharArray()){
if ((! Character.isLetter(res)) || (res != '$')){
throw new IllegalArgumentException("Only Characters and symbols allowed");
}
}
String result = discountCode.toUpperCase();
return result;
}
public Order(String itemName, int priceInCents) {
this.itemName = itemName;
this.priceInCents = priceInCents;
}
public String getItemName() {
return itemName;
}
public int getPriceInCents() {
return priceInCents;
}
public String getDiscountCode() {
return discountCode;
}
public void applyDiscountCode(String discountCode) {
this.discountCode = normalizeDiscountCode(discountCode);
}
}
public class Example {
public static void main(String[] args) {
// This is here just for example use cases.
Order order = new Order(
"Yoda PEZ Dispenser",
600);
// These are valid. They are letters and the $ character only
order.applyDiscountCode("abc");
order.getDiscountCode(); // ABC
order.applyDiscountCode("$ale");
order.getDiscountCode(); // $ALE
try {
// This will throw an exception because it contains numbers
order.applyDiscountCode("ABC123");
} catch (IllegalArgumentException iae) {
System.out.println(iae.getMessage()); // Prints "Invalid discount code"
}
try {
// This will throw as well, because it contains a symbol.
order.applyDiscountCode("w@w");
}catch (IllegalArgumentException iae) {
System.out.println(iae.getMessage()); // Prints "Invalid discount code"
}
}
}
1 Answer
andren
28,558 PointsYour if statement states that if the char
is not a letter or it is not the $ symbol then it should throw an exception. Let's go though what that will actually result in with some sample input.
Let's say the char is the letter A
. A
is a letter, so the first condition is false, but A
is not the $ symbol, so the second condition is true. With the || (OR) operator if either condition is true it will return true. So in this case an exception will be thrown. Even though A
is a valid letter. The same happens for $, while it is the $ symbol it is not a letter so one of the conditions is still true.
Basically since a character cannot be both a letter and the $ symbol at the same time there is nothing that will actually pass through your if statement without throwing an exception.
This issue can be solved in a few ways, but the simplest is just to change the || (OR) operator into the && (AND) operator. That will only return true if both of the conditions are true, which is the behavior you actually want.
Rexford Aboagye
3,080 PointsRexford Aboagye
3,080 PointsThanks for the timely response. It works also once i take out '!' from my second condition and still keep OR. Thanks again