drawing circle in directx

lcy · 2004-07-30T02:33:02

how to draw circle in directx ?

Graphics and GPU Programming Programming

Started by lcy July 27, 2004 04:49 AM

13 comments, last by nhatkthanh 19 years, 8 months ago

Mushu

1,396

July 29, 2004 04:40 PM

Quote:Original post by OrcishCoder
Another question to all: How to draw Arc and Eplise in DX8 or above?
Thanks

An arc is a portion of a circle or elipse. An elipse is a circle with the radius biased in either the x- or y-axis.

To draw an elipse, bias like this:

pkVertices [iVertex].m_fX = (float) ((float)iCenterX +((float)iRadius * xBias * cos (6.2831f*fComplete)));pkVertices [iVertex].m_fY = (float) ((float)iCenterY +((float)iRadius * yBias * sin (6.2831f*fComplete)));

To draw an arc, only calculate vertices in the area you need (ie, calculate positions in the loop for only a subset of those in the circle)

Or you could do everything my way and precalculate everything so you don't have to worry about the math

nhatkthanh

334

July 30, 2004 02:18 AM

look up for Bresenham's circle algorithm. You basically draw 1/4 of the circle and flip it 3 more times to complete the circle. This is the code that I wrote a while back for a class, just replace the drawpixel routine to a specific API to plot a pixel for a given screen coordinate.

attempting to post code.

[SOURCE]//Bresenham's circle algorithmvoid drawCircle (){	int tx,ty;	int xp = 0;	int yp = rad;	//swap(centerX,centerY);	double d = 1-rad;	while(xp <= yp)	{				tx = xp+centerX;		ty = yp+centerY;        		//second octant		drawPixle(tx,ty,1);		//third octant		drawPixle(xp-2*xp+centerX,ty,1);		//7th octant		drawPixle(tx,yp-2*(yp)+centerY,1);		//first octant		drawPixle(yp+centerX,xp+centerY,1);		//8th octant		drawPixle(yp+centerX,xp-2*xp+centerY,1);		//sixth octant		drawPixle(-1*xp+centerX,-yp+centerY,1);		//4th octant		drawPixle(-yp+centerX,(-xp+2*xp)+centerY,1);		//5th octant		drawPixle(-yp+centerX,-xp+centerY,1);		//go E		if(d < 0.0)		{			d = d + 2*xp + 3;			xp = xp+1;		}		else	//go SE		{			d = d + 2*xp - 2*yp + 5;			xp = xp+1;			yp = yp - 1;		}	}}[/SOURCE]