比较时出现问题，单击时会使点变色

Question

I'm having trouble with making my comparisons. The project is supposed to draw a square from user input, and then wherever the user clicks would draw a dot of varying colors there. For instance, if they click inside the square it should make a red circle, if it's on the edge of the square it makes a green circle, and if outside it makes a blue circle. At the moment my program draws red and blue circles, but no greens. In fact it draws red circles when it's above a certain point as well.

public class Square extends GraphicsProgram {
    // Instance variables
    private int side; // the length of a side
    private int anchorX; // the X value at the upper left corner
    private int anchorY; // the Y value at the upper left corner

    public Square(int x, int y, int side) {
        anchorX = x;
        anchorY = y;
        this.side = side;
    }

    // mouseClicked method
    public void mouseClicked(MouseEvent e) {
        // Find the location where the mouse was clicked
        int x = e.getX();
        int y = e.getY();

        // boolean variables to indicate location
        boolean isInside = false;
        boolean isOutside = false;
        boolean isOnEdge = false;

        if (x > anchorX + 1 && anchorY + 1 < y && anchorY + side + 1 > y) {
            isInside = true;
        }

        if (x > anchorX + side + 1 && anchorY + side + 1 < y && x > anchorX + side - 1 & y > anchorY + side - 1) {
            isOutside = true;
        }

        /*** NOTE: There a hard, and an easy way to do this! ***/

        if (anchorX - 1 <= x && x <= anchorX - 3 && anchorY - 1 <= y && anchorY + side - 3 >= y) {
            isOnEdge = true;
        }

        if (isOnEdge == true) {
            System.out.println("(" + x + ", " + y + ") is on the square");
            GOval circle = new GOval(x - 2, y - 2, 4, 4);
            circle.setFillColor(Color.GREEN);
            circle.setFilled(true);
            add(circle);
        }

        else if (isInside == true) {
            System.out.println("(" + x + ", " + y + ") is inside the square");
            GOval circle = new GOval(x - 2, y - 2, 4, 4);
            circle.setFillColor(Color.RED);
            circle.setFilled(true);
            add(circle);
        }

        else if (isOutside == true) {
            System.out.println("(" + x + ", " + y + ") is outside the square");
            GOval circle = new GOval(x - 2, y - 2, 4, 4);
            circle.setFillColor(Color.BLUE);
            circle.setFilled(true);
            add(circle);
        }
    }
}

We were given a hint on how to do the (x,y) locations of the square as

"For example, the left edge of the square has:

x values in the range: anchorX-1 ≤ x ≤ anchorX+1, and y values in the range: anchorY-1 ≤ y ≤ anchorY+side+1.

Which would mean that if we had a square with anchorX 50, anchorY 100 and side 60, coordinates like (49-51, 99-161) would be considered on the edge of the left side.

Answer 1

I think the problem is a little confusion with the bounds that make up the square - understandable as it's difficult to visualize.

Is Inside:

if (x > anchorX+1 && anchorY+1 < y && anchorY+side+1 > y) {
    isInside = true;
}

Right now this:

• Makes sure it's to the right of the left side x > anchorX+1
• Makes sure it's to the below the top anchorY+1 < y
• Makes sure it's above the bottom anchorY+side+1 > y

The problems with that are:

1. anchorY+side+1 > y will include the edge - the +1 will go over
2. There is nothing making sure it's to the left of the right side

A working solution would be:

if (x > anchorX+1 && y > anchorY+1 && x < anchorX+side-1 && y < anchorY+side-1) {
    isInside = true;
}

Is Outside:

if (x > anchorX+side+1 && anchorY+side+1 < y && x > anchorX+side-1 & y > anchorY+side-1) {
    isOutside = true;
}

Right now this:

• Makes sure it's to the right of the right side x > anchorX+side+1
• Makes sure it's below the bottom anchorY+side+1 < y
• Makes sure it's to the right of the right side x > anchorX+side-1
• Makes sure it's above the bottom y > anchorY+side-1

The problems with that are:

1. x > anchorX+side-1 and y > anchorY+side-1 will include the edge in the area considered outside.
2. There is nothing making sure it's to the left of the left side or above the top side.
3. x > anchorX+side+1 and x > anchorX+side-1 do practically the same.
4. You used the & symbol I'm not sure was intentional or not. The difference is && only runs if the previous are all true, while & always runs. && is therefore faster and usually preferred.
5. By checking all are true, it only runs when both to the right of and below the square.

A working solution would be:

if (x < anchorX-1 || y < anchorY-1 || x > anchorX+side+1 || y > anchorY+side+1) {
    isOutside = true;
}

Is On Edge:

This is the more tricky one

if (anchorX-1 <= x && x <= anchorX-3 && anchorY-1 <= y && anchorY+side-3 >= y) {
    isOnEdge = true;
} 

Right now this:

• Makes sure it's to the right of the left part of the edge anchorX-1 <= x
• Makes sure it's to the left of the a point to the left of the left edge x <= anchorX-3
• Makes sure it's below the top part of the bottom anchorY-1 <= y
• Makes sure it's above a point above the bottom anchorY+side-3 >= y

The problems with that are:

1. anchorX-1 <= x and x <= anchorX-3 are contridictary, so there is no x such that both are true, so the whole thing must be false.
2. You used -3 a lot, when you probably meant to use +1 to get the other side of the edge.
3. You only check the left and bottom edges - the other two are ignored.
4. As you're using && instead of || (OR), you it will only be true when the whole thing is true - it must be within both edges at the same time - meaning only the corner would be counted.

This could be fixed to create a working solution for this one, but defining all the conditions manually would take a lot of work. This is what it refers to in the easy and hard comment.

As a hint, you can use the fact you already know if it's in the square to simplify it. I suggest thinking on it.

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If you're struggling to comprehend all this, I suggest drawing it out manually and writing down all the values onto it - it's much easier than trying to do it in your head. It gets easier the more you do it.