I don't understand much of what you're doing or working with, but it looks like, without changing the logic of how you wanna paint it, you could use a for loops instead of repeating 8,16,24...64 you could make it 8 * i where i = 1,2,3...8

I think you could immediately shorten that to:

// This if statement might seem kind of magical
if (((rects.size() / 8) % 2) == 0) // if <= 16, 32, 48, 64
{
if(j%2 == 1 )
{
(*rects[(k)]).set_fill_color(Color::red);
}
else if(j%2 == 0)
{
(*rects[(k)]).set_fill_color(Color::white);
}
}
else // <= 8, 24, 40, 56
{
if(j%2 == 0 )
{
(*rects[(k)]).set_fill_color(Color::red);
}
else if(j%2 == 1)
{
(*rects[(k)]).set_fill_color(Color::white);
}
}


However, I wouldn't actually recommend the above, and if I were you, I'd do something more like:

for (int i = 0; i < rects.size(); ++i)
{
if ((i % 2) == 0)
{
rects->set_fill_color(Color::red);
}
else
{
rects->set_fill_color(Color::white);
}
}


First step is to obtain the 2-dimensional coordinate of a square. For example, if k is what you call the current square is a 1-dimensional sequential index (for example you number the squares sequentially along a row, and column by column), then you can obtain the (x,y) coordinate as x=k%8 and y=k/8.

Once you have a 2-dimensional coordinate into the checker board, you can color it according to:

if(x%2 == y%2)
red square
else
white square