Forcing a square to fin on a grid (JAVA)

[CreepyPro]

The method is supposed to draw a square on a 10x10 grid using the length and x,y coordinate an user inputs and adjust the length in case the square doesn't fit. I'm having trouble figuring out the math for adjusting the length to fit the grid.

For example, if the user inputs (x-7, y-2, length-4) my square goes out of the grid by 1.

 

public static void drawSquare(int x, int y, int len) {
        if(x+len>10)
            len = Math.max(x, len)-Math.min(x, len);
        if(y+len>10)
            len = Math.max(y, len)-Math.min(y, len);
        System.out.println("side length = " + len + ", area = " + len*len);
        drawLine(x, y, x+len, y);
        drawLine(x+len,y,x+len,y-len);
        drawLine(x+len, y-len, x, y-len);
        drawLine(x, y-len, x, y);
    }

 

[keskparane]

You are not clearly testing for any boundary conditions.

 

You could maybe compute each corner and check it's within bounds.

 

You could work off a central point of the shape. Doing this means the edge points would be a certain distance, from that central point, on each axis.

 

Should you even try to draw anything if it is outside the bounds?

 

Just some ideas and thoughts for you.

 

[Franck]

You max condition is not good at all. Should be something like :

public static void drawSquare(int x, int y, int len) {
        if(x+len>10)
            len = 10-x;
        if(y+len>10)
            len = 10-y; 
        System.out.println("side length = " + len + ", area = " + len*len);
        drawLine(x, y, x+len, y);
        drawLine(x+len,y,x+len,y-len);
        drawLine(x+len, y-len, x, y-len);
        drawLine(x, y-len, x, y);
    }

Possibly need an extra -1 to each but that will depend if you grid starts at index 0 or index 1

[KNG_HOLDY]

i dont know if you programm java gui like u programm android layout

if so just wrap it in any kind of component and make that component fill_parent