Why does my switch case work here, but if does not?

Question

I am learning how to use Lists, and in my following example the switch case works but (what I deem as) the equivalent if statement does not. Can you tell me why?

public class Kapitel14 {

public static void main(String[] args) {

    ArrayList<String> testList = new ArrayList<String>();
    testList.add("Cousin");
    testList.add("Doof");
    testList.add("Dorf");
    testList.add("Dortmund");
    testList.add("Franz");
    System.out.println(listCount(testList));
}

public static int listCount(ArrayList<String> newList) {

    int capDCounter = 0;
    for (String element : newList) {
        String firstLetter = Character.toString(element.charAt(0));
        switch (firstLetter) {
            case ("D"):
                capDCounter++;
                break;

            default:
                continue;
        }

        //if I use this instead it returns wrong results:
        //if (firstLetter == "D") 
        //  capDCounter++;
    }
    return capDCounter;
}

You can just replace "D" by 'D'. – daniu Apr 20 '18 at 12:09 — daniu, Apr 20 '18 at 12:09

score 1 · Accepted Answer · answered Apr 20 '18 at 12:00

1

Use

if (firstLetter.equals("D"))
             capDCounter++;

instead of

if (firstLetter == "D") 
          capDCounter++;

.equals() method should be used here as you want to compare the values of strings.

answered Apr 20 '18 at 12:00

Shanu Gupta

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Thank you! Is it because my version would have compared the adress of firstLetter to "D"? – Crenshaw Apr 20 '18 at 12:03
Apparently yes. But I would term them as references instead of address. BTW thats close analogy to C/C++. – Shanu Gupta Apr 20 '18 at 12:06

Why does my switch case work here, but if does not?

1 Answers1