String equal in Java

Question

I have to write a hangman homework assignment. We aren't allowed to use any char array and only use String methods to manipulate the string itself. A problem I'm running into is when I try to check the string builder for duplicates.

This is my hangman method

public void playHangMan(Scanner guessWord)
{
    int error = 6;
    String letter;
    boolean vali, dup;

    displayHangman();// call displayHangman method

    // get input, lower case input, and then validate in a loop
    do
    {
        letter = guessWord.nextLine();
        letter = lowerGuess(letter); // call lowerGuess method
        vali = letter.matches("[a-z]");

        dup = checkDup(letter); // call checkDup method

        if(vali == false)
        {
            System.out.print("Please enter only a SINGLE letter: ");
        }
        else if(dup == true)
        {
            System.out.print("Duplicated letter, please enter another: ");
        }

        if(dup == false && vali == true)
        {
            this.guessAns.append(letter);
        }

    }
    while(vali == false && dup == false);


}// end playHangman method

My duplicate method:

private boolean checkDup(String letter)
{
    int i;
    boolean dup = false;

    // check guessAns StringBuilder for duplicate letters
    for(i = 0; i <= this.guessAns.length() - 1 && dup == false; i++)
    {
        if(letter.equals(this.guessAns.charAt(i)))
        {
            dup = true;
        }
    }

    if(dup == true)
    {
        return true;
    }
    else
    {
        return false;
    }   
}// end checkDup method

The problem is that my checkDup method isn't finding any duplicates. I tried appending the letter a into my string builder and entering in the value a for letter, but still no luck. Can it be that letter.equals(this.guessAns.charAt(i)) is comparing a String to a Char and that's why my checkDup method is failing to find duplicates? Can someone explain a way that I can get around this? Any help will be greatly appreciated.

Note that the bottom half of checkDup can be simplified to return dup;. — Paul Bellora
– Paul Bellora, Commented Nov 3, 2013 at 4:42
@PaulBellora .. which can itself be simplified by removing the dup variable. — user2864740
– user2864740, Commented Nov 3, 2013 at 4:42
@PaulBellora Thanks! You're right. Will definitely change that up. — Nathan
– Nathan, Commented Nov 3, 2013 at 4:43
@user2864740 has a point - you could replace dup = true; with return true; and put return false; at the end. +1 for a well-written question btw. — Paul Bellora
– Paul Bellora, Commented Nov 3, 2013 at 4:44

Josh M · Accepted Answer · 2013-11-03 04:30:04Z

4

2 ways you could work around this:

letter.charAt(0) == guessAns.charAt(i);

The solution above is under the assumption that letter is a single char.

The other way is to invoke Character.toString on the char:

letter.equals(Character.toString(guessAns.charAt(i)));

answered Nov 3, 2013 at 4:30

Josh M

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1 Comment

Nathan Over a year ago

Thanks for the quick reply. Your input really helped out.

Brigham · Accepted Answer · 2013-11-03 04:30:20Z

4

Calling String.equals(char) will always return false, as Strings never compare as equal to Characters. Do this to compare them as characters:

if(letter.charAt(0) == this.guessAns.charAt(i)) {

answered Nov 3, 2013 at 4:30

Brigham

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Comments

Justin · Accepted Answer · 2013-11-03 05:46:49Z

There is another way of checking for duplicates: using a regex. The regex .*([a-z])\1.* will match any String with a repeated character between a and z. To go more general, .*(.)\1.* will match any String with any repeated character.

dup = letter.matches(".*([a-z])\\1.*");

How this works: . matches any character and* means 0 or more times, so .* means any amount of any character (it "consumes" any amount of characters). ( starts a group; ) ends a group. [a-z] matches any character between a to z. \1 is a backreference. Therefore, the regex looks for any character [a-z] which is immediately followed by the same character.

Note: If you are looking to see if the String contains the same character even when separated, then this should work: .*([a-z]).*\1.*

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