StringBuffer - append using charAt returns null

Question

I can't seem to find the answer to this - apologies if this question has already been asked.

In the Java code below I'm trying to write a method that returns the numeric values (as a string) that would have been pressed on a mobile keypad to generate the input String e.g. "cab" -> 222.

The HashMap isn't empty, however following the append to the StringBuffer object sbf, when I print wordToSignature("cab") in main, it returns nullnullnull.

The problem appears to be with the word.charAt(i) code. When I test the append without word.charAt(i), it performs the append and the value in the HashMap is printed.

I would be very grateful if someone could indicate where I might be going wrong.

Thanks in advance.

public static String wordToSignature(String word) {

    HashMap <String, String> keypad = new HashMap <String, String>();

    keypad.put("a", "2"); keypad.put("b", "2"); keypad.put("c", "2");
    keypad.put("d", "3"); keypad.put("e", "3"); keypad.put("f", "3");
    keypad.put("g", "4"); keypad.put("h", "4"); keypad.put("i", "4");
    keypad.put("j", "5"); keypad.put("k", "5"); keypad.put("l", "5");
    keypad.put("m", "6"); keypad.put("n", "6"); keypad.put("o", "6");
    keypad.put("p", "7"); keypad.put("q", "7"); keypad.put("r", "7"); keypad.put("s", "7");
    keypad.put("t", "8"); keypad.put("u", "8"); keypad.put("v", "8");
    keypad.put("w", "9");   keypad.put("x", "9");   keypad.put("y", "9"); keypad.put("z", "9");

    StringBuffer sbf = new StringBuffer("");

    for(int i = 0; i < word.length(); i++) {
        sbf.append(keypad.get(word.charAt(i)));
    }
    return sbf.toString();
}

^ String#charAt(int) returns a char. You've stored String keys in your map. — Sotirios Delimanolis
– Sotirios Delimanolis, Commented Jan 29, 2015 at 2:32

Óscar López · Accepted Answer · 2015-01-29 02:34:46Z

1

Try this:

HashMap<Character, String> keypad = new HashMap<Character, String>();
keypad.put('a', "2"); keypad.put('b', "2"); keypad.put('c', "2");
// and so on - notice the single quotes!

If you're going to iterate over the characters of a String, then you might as well store Characters and not Strings as keys in the Map.

answered Jan 29, 2015 at 2:34

Óscar López

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3 Comments

user3742439 Over a year ago

Thank you very much, that has solved the problem. Sorry, I am a beginner, but what was the problem?

Óscar López Over a year ago

@user3742439 if this answer solved your problem, please consider accepting it, just click on the check mark to its left ;)

Óscar López Over a year ago

@user3742439 the problem is that when you call word.charAt(i) it returns a char, but the map was using a String as a key. Both types must be consistent, the easiest way to fix this was to also use char as a key.

Joe · Accepted Answer · 2015-01-29 02:36:48Z

Try the following, instead of using a string in the keypad hashmap use a Char.

public static String wordToSignature(String word) {

    HashMap <Character, String> keypad = new HashMap <Character, String>();

    keypad.put('a', "2"); keypad.put('b', "2"); keypad.put('c', "2");
    keypad.put('d', "3"); keypad.put('e', "3"); keypad.put('f', "3");
    keypad.put('g', "4"); keypad.put('h', "4"); keypad.put('i', "4");
    keypad.put('j', "5"); keypad.put('k', "5"); keypad.put('l', "5");
    keypad.put('m', "6"); keypad.put('n', "6"); keypad.put('o', "6");
    keypad.put('p', "7"); keypad.put('q', "7"); keypad.put('r', "7"); keypad.put('s', "7");
    keypad.put('t', "8"); keypad.put('u', "8"); keypad.put('v', "8");
    keypad.put('w', "9");   keypad.put('x', "9");   keypad.put('y', "9"); keypad.put('z', "9");

    StringBuffer sbf = new StringBuffer("");

    for(int i = 0; i < word.length(); i++) {
        sbf.append(keypad.get(word.charAt(i)));
    }
    return sbf.toString();
}

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