Cohen–Sutherland algorithm

Question

I was reading Cohen–Sutherland algorithm on Wikipedia and I am confused in these two parts,

int ComputeOutCode(double x, double y) {
    ...
    if (x < xmin)       code |= LEFT;
    else if (x > xmax)  code |= RIGHT;
    if (y < ymin)       code |= BOTTOM;
    else if (y > ymax)  code |= TOP;
    ...
}

and

if (outcodeOut & TOP) {           // point is above the clip rectangle
    x = x0 + (x1 - x0) * (ymax - y0) / (y1 - y0);
    y = ymax;
} else if (outcodeOut & BOTTOM) { // point is below the clip rectangle
    x = x0 + (x1 - x0) * (ymin - y0) / (y1 - y0);
    y = ymin;
} else if (outcodeOut & RIGHT) {  // point is to the right of clip rectangle
    y = y0 + (y1 - y0) * (xmax - x0) / (x1 - x0);
    x = xmax;
} else if (outcodeOut & LEFT) {   // point is to the left of clip rectangle
    y = y0 + (y1 - y0) * (xmin - x0) / (x1 - x0);
    x = xmin;
}

What I am confused is, How does it handle points who are top-left, top right, etc. It just handles top, left, right and bottom.

Why use | in code |= LEFT; and outcodeOut & TOP in if-else below it? I know that these two somehow handle all 8 possible cases but I don't know how.

Answer 1

code is a bit field which uses bitwise boolean operators to store whether certain flags are set or not.

code |= LEFT; is short for code = code | LEFT; and sets the LEFT flag. outcodeOut & LEFT is an expression that is only true (!=0) when the LEFT flag is set.

The shown excerpt does indeed not handle all (16) possible cases at once, but it's part of a larger loop that may run multiple times. So it's done in several iterations.