Java: Accessing pointers

Question

I get: 'unexpected type
required: variable
found : value' on the marked lines (*)

for (int i = 0; i < boardSize; i++) {
    for (int j = 0; j < boardSize; j++) {
        rows[i].getSquare(j) = matrix[i][j]; // * points to the ( in (j)
        columns[j].getSquare(i) = matrix[i][j]; // * points to the ( in
        int[] b = getBox(i, j);
        int[] boxCord = getBoxCoordinates(i + 1, j + 1);
        boxes[b[0]][b[1]].getSquare(boxCord[0], boxCord[1]);
    }
}

This is my Row class:

private Square[] row;

Row(int rowCount) {
    this.row = new Square[rowCount];
}

public Square getSquare(int index) {
    return this.row[index];
}

Please help me out by pointing out what I'm doing wrong here.
Thanks in advance.

There are some good answers on your previous questions, you really should accept some of them — Perception, Apr 09 '12 at 15:35

score 10 · Answer 1 · answered Apr 09 '12 at 15:34

You cannot assign something to the return value of a method. Instead, you need to add a setSquare() method to the Row class:

public Square setSquare(int index, Square value) {
    this.row[index] = value;
}

and use it like this:

rows[i].setSquare(j, matrix[i][j]);

score 0 · Answer 2 · answered Apr 09 '12 at 15:37

Java doesn't have pointers - references are not the same thing.

It's impossible to tell what's really going on based on the code you posted. I think you need a method to set the value of that Square in the private array your Row class owns.

public void setSquare(int index, Square newSquare) {
    this.row[index] = newSquare; 
}

It looks like a poor abstraction, in any case.

score -2 · Answer 3 · answered Apr 09 '12 at 15:33

-2

Java has no pointers. Objects are passed and returned by reference. In terms of C++, rows[i].getSquare(j) is an rvalue, not an lvalue, so you cannot assign to it.

Instead, you should create and use rows[i].setSquare(...).

answered Apr 09 '12 at 15:33

Alexander Pavlov

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Objects aren't passed by reference. References are passed by value. – Jon Skeet Apr 09 '12 at 15:34
The "pass by reference" thing is flaky. But on the other hand the lvalue/rvalue stuff is right. – A.H. Apr 09 '12 at 15:36
Right you are, sorry for the awkward phrasing. – Alexander Pavlov Apr 09 '12 at 15:36
See my answer for further explanation on how Java passes references by value: http://stackoverflow.com/a/9404727/597657 – Eng.Fouad Apr 09 '12 at 15:37

Java: Accessing pointers

3 Answers3