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Juan Rodríguez
782 PointsWhy can't I return the counter in this excercise?
I can't seem to return the counter. The excercises asks to increment a counter if there is a tile in the player's hand in a method that is called getTileCount(). So I increment the counter but I CAN'T RETURN IT since the "return" method only returns strings. How can I finish these challenge? I'M STUCKED! In the video we returned a String so there wasn't any problem, in here we must return a count, ergo, an Integer, how can I do it?
public class ScrabblePlayer {
public static String mHand;
public ScrabblePlayer() {
mHand = "";
}
public String getHand() {
return mHand;
}
public void addTile(char tile) {
// Adds the tile to the hand of the player
mHand += tile;
}
public boolean hasTile(char tile) {
return mHand.indexOf(tile) > -1;
}
public String getTileCount()
{
int counter = 0;
for (char tileWeNeedToCount : mHand.toCharArray()){
if (mHand.indexOf(tileWeNeedToCount) >= 0) {
counter += 1;
}
}
return counter;
}
}
3 Answers
Jeremy Hill
29,567 Pointsin your for loop you need to add a type in there so it knows what type of data to expect:
for(char letter : mHand.toCharArray()) // then in the loop make sure you reference 'letter' or whatever you name it.
Jeremy Hill
29,567 PointsIt is because you set your method up to return a string:
It should be: public int getTileCount(char tile){ }
Jeremy Hill
29,567 PointsAlso, you have to give it a char parameter so the counter will know which char to count.
Juan Rodríguez
782 PointsOkay, so I set the method to return an int and I put a parameter for it to accept a char, so that we know which tile we need to count. Like this:
public int getTileCount(char tile) { int counter = 0; for (tile : mHand.toCharArray()) { if (mHand.indexOf(tile) >= 0) { counter += 1; } } return counter; } }
But now it says:
./ScrabblePlayer.java:24: error: bad initializer for for-loop for (tile : mHand.toCharArray()){ ^
So what do I do, why can't I call the char I had set up and use it for the each loop? If I declare it again in the for parameters I will say it's an already-defined variable. So what can I do?
Jeremy Hill
29,567 PointsGood deal! You're welcome :)
Juan Rodríguez
782 PointsJuan Rodríguez
782 PointsThank you a lot! :D That worked just fine.