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Chaz Hall
1,970 PointsArguments in Java?
Arguments are variables that are utilized by that object in a class only. The method uses the argument to declare things
public void addTile(char tile); mHand+=tile;
public boolean hasTile(char tile); boolean hasTile=mHand.indexOf(tile)!=1; return false;
The char tile (char tile) is the same name in two different methods. Is this the same variable throughout the class or is it unique to the method? I know that the name of each (addTile and hasTile) are different and it necessarily wouldn't matter, I wouldn't think. The only reason I would think it was the same is because of char tile.
I'm trying to understand the logic and this has just slightly tripped me up. Thanks!
public class ScrabblePlayer {
private String mHand;
public ScrabblePlayer() {
mHand = "";
}
public String getHand() {
return mHand;
}
public void addTile(char tile) {
mHand += tile;
}
public boolean hasTile(char tile) {
boolean hasTile = mHand.indexOf(tile) != -1;
return false;
}
}
2 Answers
jcorum
71,830 PointsWhen you create a method and specify an input argument, you can name that argument whatever you want (although the name should comply with the naming conventions).
So, in
public void addTile(char tile) { }
it could be (char aTile) or (char myTile) or (char inputTile) just as easily as (char tile).
What you name it has nothing to do with the name of an input (or formal) parameter of any other function.
What Java does, when it runs the function, is it creates a local variable, e.g., named tile, whose scope is the function's code block. After the function runs, that local variable goes out of scope.
iftekhar uddin
4,194 PointsJcorum pretty much summed it up but in simple terms. The variable you're talking about (the one created in the parameter) is destroyed after the method is finished and it can only be used within that method unless you specifically pass it into another function.
Ex: lets say that hasTitle had parameter (char mTile) instead of (char tile). You cannot use the variable tile in the hasTile function because it's not in the same scope.