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.

• andy256: So, slipping into coaching mode: what do you understand |= does?

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.

• Sourabh: Now I am confused how line (200,200)->(900,850) is clipped for clip rectangle (100,100) (800,800). If we use outcodeOut&RIGHT block, it'll get solved. But the algorithm will use TOP block then RIGHT block, but its still solved somehow.