Bresenham's Midpoint Algorithm, How do I fill in the gaps?

Question

I am attempting to draw a circle (and eventually fill the entire circle) using Bresenham's Midpoint Algorithm. It will become the outline for fog of war in a basic 2D game.

Instead of drawing a line or filling a pixel I am changing values in a 2D vector. This works correctly and the locations in mapMatrix that are changed in this code display correctly.

However only the bottom and top sections of the circle appear. How do I fill in the gaps?

int d = 3 - (2 * radius);
int x = 0;
int y = radius;

while (x <= y)
{
    mapMatrix[centerX + x][centerY + y].fog = false;
    mapMatrix[centerX + x][centerY - y].fog = false;
    mapMatrix[centerX - x][centerY + y].fog = false;
    mapMatrix[centerX - x][centerY - y].fog = false;
    mapMatrix[centerX + x][centerY + y].fog = false;
    mapMatrix[centerX + x][centerY - y].fog = false;
    mapMatrix[centerX - x][centerY + y].fog = false;
    mapMatrix[centerX - x][centerY - y].fog = false;

    if (d < 0) 
    {
        d = (d + (4*x) + 6);
    } 
    else 
    {
        d = ((d + 4 * (x - y)) + 10);
        y--;
    }
    x++;
} 

I can put a picture of my output so please see this crude ASCII drawing.

-------------
----ooooo----
---o-----o---
-------------
-------------
-------------
------o------
-------------
-------------
-------------
---o-----o---
----ooooo----
-------------

Thanks in advance!

Answer 1

A more complete answer:

You are looping while (x<=y). that means that the last iteration is when x == y. but x == y only on the diagonal so that is where you stop:

x------------
-x--ooooo----
--xo-----o---
---x---------
----x--------
-----x-------
------x------
-------x-----
--------x----
---------x---
---o-----ox--
----ooooo--x-
------------x

You are not iterating on the circle, you are only iterating on a line (x), and you calculate top and bottom y's.

Note that there are only two y's for every x and, near the end, you would need more y's for every x. This is why you need to iterate again but this time on a column (y) and calculate two x's for every y, basically switching x and y in the above algorithm.