Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

Java Java Objects Creating the MVP Counting Scrabble Tiles

Hai Phan
Hai Phan
2,442 Points

About the challenge

This is my first try:

public class ScrabblePlayer {
  // A String representing all of the tiles that this player has
  private String tiles;

  public ScrabblePlayer() {
    tiles = "";
  }

  public String getTiles() {
    return tiles;
  }

  public void addTile(char tile) {
    tiles += tile;
  }

  public boolean hasTile(char tile) {
    return tiles.indexOf(tile) != -1;
  }

  public int getCountOfLetter(char letter) {
    int count = 0;
    for (char check : tiles.toCharArray()) {
        if (letter == check) {
      count += 1;
      }
    }
    return count;
  }
}

and it fails. But when I replace

count += 1;

with

count ++;

I pass. I think it will returns the same value. But why count +=1 is not accept?

1 Answer

Steven Parker
Steven Parker
231,269 Points

It might be functionally equivalent, but apparently the validator is looking for that specific type of increment. The error message you get with the code above says: "Bummer: Make sure you create a variable that you can increment when you find a match during your loop. Use the ++ shortcut!"

My guess is that they just want you to practice the techniques as shown in the video.

Hai Phan
Hai Phan
2,442 Points

I thought the same, but because I'm new I just want to make sure is there anything else that I missed :). Thank you for the answer anyway.