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I'm currently following Andre Lamothe's book Tricks of the Windows Game Programming Gurus and trying to write my own software renderer with Win32 GDI. However, I got stuck at the triangle rasterization part of the book where the author is using Bresenham's line algorithm. I managed to get the rasterization to work on DirectX, but not on Win32 GDI despite identical rendering code. Both renders a triangle but the DirectX one renders it correctly while the Win32 almost renders it correctly
Link to my code: https://pastebin.com/VwySmQXS
It could be worth mention that the DirectX version is written in c++, while my own is written in C. I wasn't able to get a good screenshot of the two triangles so here is a picture painted in paint of how they both look.
Left is DirectX (how I want it to look), right is Win32 GDI:
Am I missing something here? Both draw_triangle_2d are called with the same arguments.
static void draw_top_tri(struct render_buffer *buffer,
int x1, int y1,
int x2, int y2,
int x3, int y3,
float r, float g, float b, float a)
{
unsigned int color = vec4_to_uint(vec4_mul(Vec4(r, g, b, a), 255.0f));
// this function draws a triangle that has a flat top
float dx_right, // the dx/dy ratio of the right edge of line
dx_left, // the dx/dy ratio of the left edge of line
xs, xe, // the starting and ending points of the edges
height; // the height of the triangle
int temp_x, // used during sorting as temps
temp_y,
right, // used by clipping
left;
// destination address of next scanline
unsigned int *dest_addr = NULL;
// test order of x1 and x2
if(x2 < x1)
{
temp_x = x2;
x2 = x1;
x1 = temp_x;
} // end if swap
// compute delta's
height = (float)(y3 - y1);
dx_left = (x3 - x1) / height;
dx_right = (x3 - x2) / height;
// set starting points
xs = (float)x1;
xe = (float)x2 + (float)0.5;
// perform y clipping
if(y1 < MIN_CLIP_Y)
{
// compute new xs and ys
xs = xs + dx_left * (float)(-y1 + MIN_CLIP_Y);
xe = xe + dx_right * (float)(-y1 + MIN_CLIP_Y);
// reset y1
y1 = MIN_CLIP_Y;
} // end if top is off screen
if(y3 > MAX_CLIP_X)
y3 = MAX_CLIP_X;
// compute starting address in video memory
dest_addr = (unsigned int *)buffer->memory + y1 * buffer->width;
// test if x clipping is needed
if(x1 >= MIN_CLIP_X && x1 <= MAX_CLIP_X &&
x2 >= MIN_CLIP_X && x2 <= MAX_CLIP_X &&
x3 >= MIN_CLIP_X && x3 <= MAX_CLIP_X)
{
// draw the triangle
for(temp_y = y1; temp_y <= y3; temp_y++, dest_addr += buffer->width)
{
memset((unsigned int *)dest_addr + (unsigned int)xs,
color, (unsigned int)(xe - xs + 1) * 4);
// adjust starting point and ending point
xs += dx_left;
xe += dx_right;
} // end for
} // end if no x clipping needed
else
{
// clip x axis with slower version
// draw the triangle
for(temp_y = y1; temp_y <= y3; temp_y++, dest_addr += buffer->width)
{
// do x clip
left = (int)xs;
right = (int)xe;
// adjust starting point and ending point
xs += dx_left;
xe += dx_right;
// clip line
if(left < MIN_CLIP_X)
{
left = MIN_CLIP_X;
if(right < MIN_CLIP_X)
continue;
}
if(right > MAX_CLIP_X)
{
right = MAX_CLIP_X;
if(left > MAX_CLIP_X)
continue;
}
memset((unsigned int *)dest_addr + (unsigned int)left,
color, (unsigned int)(right - left + 1) * 4);
} // end for
} // end else x clipping needed
} // end Draw_Top_Tri
static void
draw_bottom_tri(struct render_buffer *buffer,
int x1, int y1,
int x2, int y2,
int x3, int y3,
float r, float g, float b, float a)
{
unsigned int color = vec4_to_uint(vec4_mul(Vec4(r, g, b, a), 255.0f));
// this function draws a triangle that has a flat bottom
float dx_right, // the dx/dy ratio of the right edge of line
dx_left, // the dx/dy ratio of the left edge of line
xs, xe, // the starting and ending points of the edges
height; // the height of the triangle
int temp_x, // used during sorting as temps
temp_y,
right, // used by clipping
left;
// destination address of next scanline
unsigned int *dest_addr;
// test order of x1 and x2
if(x3 < x2)
{
temp_x = x2;
x2 = x3;
x3 = temp_x;
} // end if swap
// compute delta's
height = (float)(y3 - y1);
dx_left = (x2 - x1) / height;
dx_right = (x3 - x1) / height;
// set starting points
xs = (float)x1;
xe = (float)x1; // +(float)0.5;
// perform y clipping
if(y1 < MIN_CLIP_Y)
{
// compute new xs and ys
xs = xs + dx_left * (float)(-y1 + MIN_CLIP_Y);
xe = xe + dx_right * (float)(-y1 + MIN_CLIP_Y);
// reset y1
y1 = MIN_CLIP_Y;
} // end if top is off screen
if(y3 > MAX_CLIP_X)
y3 = MAX_CLIP_X;
// compute starting address in video memory
dest_addr = (unsigned int *)buffer->memory + y1 * buffer->width;
// test if x clipping is needed
if(x1 >= MIN_CLIP_X && x1 <= MAX_CLIP_X &&
x2 >= MIN_CLIP_X && x2 <= MAX_CLIP_X &&
x3 >= MIN_CLIP_X && x3 <= MAX_CLIP_X)
{
// draw the triangle
for(temp_y = y1; temp_y <= y3; temp_y++, dest_addr += buffer->width)
{
memset((unsigned int *)dest_addr + (unsigned int)xs,
color, (unsigned int)(xe - xs + 1) * 4);
// adjust starting point and ending point
xs += dx_left;
xe += dx_right;
} // end for
} // end if no x clipping needed
else
{
// clip x axis with slower version
// draw the triangle
for(temp_y = y1; temp_y <= y3; temp_y++, dest_addr += buffer->width)
{
// do x clip
left = (int)xs;
right = (int)xe;
// adjust starting point and ending point
xs += dx_left;
xe += dx_right;
// clip line
if(left < MIN_CLIP_X)
{
left = MIN_CLIP_X;
if(right < MIN_CLIP_X)
continue;
}
if(right > MAX_CLIP_X)
{
right = MAX_CLIP_X;
if(left > MAX_CLIP_X)
continue;
}
memset((unsigned int *)dest_addr + (unsigned int)left,
color, (unsigned int)(right - left + 1) * 4);
} // end for
} // end else x clipping needed
} // end Draw_Bottom_Tri}
ENGINE_CORE_EXPORT void
draw_triangle_2d(struct render_buffer *buffer,
int x1, int y1,
int x2, int y2,
int x3, int y3,
float r, float g, float b, float a)
{
// this function draws a triangle on the destination buffer
// it decomposes all triangles into a pair of flat top, flat bottom
int temp_x, // used for sorting
temp_y,
new_x;
// test for h lines and v lines
if((x1 == x2 && x2 == x3) || (y1 == y2 && y2 == y3))
return;
// sort p1,p2,p3 in ascending y order
if(y2 < y1)
{
temp_x = x2;
temp_y = y2;
x2 = x1;
y2 = y1;
x1 = temp_x;
y1 = temp_y;
} // end if
// now we know that p1 and p2 are in order
if(y3 < y1)
{
temp_x = x3;
temp_y = y3;
x3 = x1;
y3 = y1;
x1 = temp_x;
y1 = temp_y;
} // end if
// finally test y3 against y2
if(y3 < y2)
{
temp_x = x3;
temp_y = y3;
x3 = x2;
y3 = y2;
x2 = temp_x;
y2 = temp_y;
} // end if
// do trivial rejection tests for clipping
if(y3 < MIN_CLIP_Y || y1 > MAX_CLIP_X ||
(x1 < MIN_CLIP_X && x2 < MIN_CLIP_X && x3 < MIN_CLIP_X) ||
(x1 > MAX_CLIP_X && x2 > MAX_CLIP_X && x3 > MAX_CLIP_X))
return;
// test if top of triangle is flat
if(y1 == y2)
{
draw_top_tri(buffer, x1, y1, x2, y2, x3, y3, r, g, b, a);
} // end if
else
if(y2 == y3)
{
draw_bottom_tri(buffer, x1, y1, x2, y2, x3, y3, r, g, b, a);
} // end if bottom is flat
else
{
// general triangle that's needs to be broken up along long edge
new_x = x1 + (int)(0.5 + (float)(y2 - y1) * (float)(x3 - x1) / (float)(y3 - y1));
// draw each sub-triangle
draw_bottom_tri(buffer, x1, y1, new_x, y2, x2, y2, r, g, b, a);
draw_top_tri(buffer, x2, y2, new_x, y2, x3, y3, r, g, b, a);
} // end else
} // end Draw_Triangle_2D
I'm trying to draw a flat bottom triangle, but its drawn as flat top triangle.
I would like to know what I'm doing wrong, in terms of math
here is my code
void Rasterizer::DrawBottomTriangle(int x0, int y0, int x1, int y1, int x2, int y2, Color color)
{
int temp_x;
// test order of x1 and x2
if (x2 < x1)
{
temp_x = x1;
x1 = x2;
x2 = temp_x;
} //
float dxy_left = (float)(x2 - x0) / (y2 - y0);
float dxy_right = (float)(x1 - x0) / (y1 - y0);
// set starting and ending points for edge trace
float xs = x0;
float xe = x0;
// draw each scanline
for (int y = y0; y >= y1; y--)
{
// draw a line from xs to xe at y in color c
DrawLine(color, (int)xs, y, color, (int)xe, (int)y);
// move down one scanline
xs += dxy_left;
xe += dxy_right;
} // end for y
}
I'm trying to implement the Delaunay triangulation in C++. Currently it's working, but I'm not getting the correct amount of triangles.
I try it with 4 points in a square pattern : (0,0), (1,0), (0,1), (1,1).
Here's the algorithm I use :
std::vector<Triangle> Delaunay::triangulate(std::vector<Vec2f> &vertices) {
// Determinate the super triangle
float minX = vertices[0].getX();
float minY = vertices[0].getY();
float maxX = minX;
float maxY = minY;
for(std::size_t i = 0; i < vertices.size(); ++i) {
if (vertices[i].getX() < minX) minX = vertices[i].getX();
if (vertices[i].getY() < minY) minY = vertices[i].getY();
if (vertices[i].getX() > maxX) maxX = vertices[i].getX();
if (vertices[i].getY() > maxY) maxY = vertices[i].getY();
}
float dx = maxX - minX;
float dy = maxY - minY;
float deltaMax = std::max(dx, dy);
float midx = (minX + maxX) / 2.f;
float midy = (minY + maxY) / 2.f;
Vec2f p1(midx - 20 * deltaMax, midy - deltaMax);
Vec2f p2(midx, midy + 20 * deltaMax);
Vec2f p3(midx + 20 * deltaMax, midy - deltaMax);
// Add the super triangle vertices to the end of the vertex list
vertices.push_back(p1);
vertices.push_back(p2);
vertices.push_back(p3);
// Add the super triangle to the triangle list
std::vector<Triangle> triangleList = {Triangle(p1, p2, p3)};
// For each point in the vertex list
for(auto point = begin(vertices); point != end(vertices); point++)
{
// Initialize the edges buffer
std::vector<Edge> edgesBuff;
// For each triangles currently in the triangle list
for(auto triangle = begin(triangleList); triangle != end(triangleList);)
{
if(triangle->inCircumCircle(*point))
{
Edge tmp[3] = {triangle->getE1(), triangle->getE2(), triangle->getE3()};
edgesBuff.insert(end(edgesBuff), tmp, tmp + 3);
triangle = triangleList.erase(triangle);
}
else
{
triangle++;
}
}
// Delete all doubly specified edges from the edge buffer
// Black magic by https://github.com/MechaRage
auto ite = begin(edgesBuff), last = end(edgesBuff);
while(ite != last) {
// Search for at least one duplicate of the current element
auto twin = std::find(ite + 1, last, *ite);
if(twin != last)
// If one is found, push them all to the end.
last = std::partition(ite, last, [&ite](auto const &o){ return !(o == *ite); });
else
++ite;
}
// Remove all the duplicates, which have been shoved past "last".
edgesBuff.erase(last, end(edgesBuff));
// Add the triangle to the list
for(auto edge = begin(edgesBuff); edge != end(edgesBuff); edge++)
triangleList.push_back(Triangle(edge->getP1(), edge->getP2(), *point));
}
// Remove any triangles from the triangle list that use the supertriangle vertices
triangleList.erase(std::remove_if(begin(triangleList), end(triangleList), [p1, p2, p3](auto t){
return t.containsVertex(p1) || t.containsVertex(p2) || t.containsVertex(p3);
}), end(triangleList));
return triangleList;
}
And here's what I obtain :
Triangle:
Point x: 1 y: 0
Point x: 0 y: 0
Point x: 1 y: 1
Triangle:
Point x: 1 y: 0
Point x: 1 y: 1
Point x: 0 y: 1
Triangle:
Point x: 0 y: 0
Point x: 1 y: 1
Point x: 0 y: 1
While this would be the correct output :
Triangle:
Point x: 1 y: 0
Point x: 0 y: 0
Point x: 0 y: 1
Triangle:
Point x: 1 y: 0
Point x: 1 y: 1
Point x: 0 y: 1
I have no idea why there is a triangle with the (0, 0) and the (1, 1).
I need an outside eye to review the code and find out what's going wrong.
All the sources are on my Github repo. Feel free to fork it and to PR your code.
Thanks!
what about this implementation of Paul Bourke's Delaunay triangulation algorithm. Take a look at Triangulate() I have used this source many times without any complains
#include <iostream>
#include <stdlib.h> // for C qsort
#include <cmath>
#include <time.h> // for random
const int MaxVertices = 500;
const int MaxTriangles = 1000;
//const int n_MaxPoints = 10; // for the test programm
const double EPSILON = 0.000001;
struct ITRIANGLE{
int p1, p2, p3;
};
struct IEDGE{
int p1, p2;
};
struct XYZ{
double x, y, z;
};
int XYZCompare(const void *v1, const void *v2);
int Triangulate(int nv, XYZ pxyz[], ITRIANGLE v[], int &ntri);
int CircumCircle(double, double, double, double, double, double, double, double, double&, double&, double&);
using namespace std;
////////////////////////////////////////////////////////////////////////
// CircumCircle() :
// Return true if a point (xp,yp) is inside the circumcircle made up
// of the points (x1,y1), (x2,y2), (x3,y3)
// The circumcircle centre is returned in (xc,yc) and the radius r
// Note : A point on the edge is inside the circumcircle
////////////////////////////////////////////////////////////////////////
int CircumCircle(double xp, double yp, double x1, double y1, double x2,
double y2, double x3, double y3, double &xc, double &yc, double &r){
double m1, m2, mx1, mx2, my1, my2;
double dx, dy, rsqr, drsqr;
/* Check for coincident points */
if(abs(y1 - y2) < EPSILON && abs(y2 - y3) < EPSILON)
return(false);
if(abs(y2-y1) < EPSILON){
m2 = - (x3 - x2) / (y3 - y2);
mx2 = (x2 + x3) / 2.0;
my2 = (y2 + y3) / 2.0;
xc = (x2 + x1) / 2.0;
yc = m2 * (xc - mx2) + my2;
}else if(abs(y3 - y2) < EPSILON){
m1 = - (x2 - x1) / (y2 - y1);
mx1 = (x1 + x2) / 2.0;
my1 = (y1 + y2) / 2.0;
xc = (x3 + x2) / 2.0;
yc = m1 * (xc - mx1) + my1;
}else{
m1 = - (x2 - x1) / (y2 - y1);
m2 = - (x3 - x2) / (y3 - y2);
mx1 = (x1 + x2) / 2.0;
mx2 = (x2 + x3) / 2.0;
my1 = (y1 + y2) / 2.0;
my2 = (y2 + y3) / 2.0;
xc = (m1 * mx1 - m2 * mx2 + my2 - my1) / (m1 - m2);
yc = m1 * (xc - mx1) + my1;
}
dx = x2 - xc;
dy = y2 - yc;
rsqr = dx * dx + dy * dy;
r = sqrt(rsqr);
dx = xp - xc;
dy = yp - yc;
drsqr = dx * dx + dy * dy;
return((drsqr <= rsqr) ? true : false);
}
///////////////////////////////////////////////////////////////////////////////
// Triangulate() :
// Triangulation subroutine
// Takes as input NV vertices in array pxyz
// Returned is a list of ntri triangular faces in the array v
// These triangles are arranged in a consistent clockwise order.
// The triangle array 'v' should be malloced to 3 * nv
// The vertex array pxyz must be big enough to hold 3 more points
// The vertex array must be sorted in increasing x values say
//
// qsort(p,nv,sizeof(XYZ),XYZCompare);
///////////////////////////////////////////////////////////////////////////////
int Triangulate(int nv, XYZ pxyz[], ITRIANGLE v[], int &ntri){
int *complete = NULL;
IEDGE *edges = NULL;
IEDGE *p_EdgeTemp;
int nedge = 0;
int trimax, emax = 200;
int status = 0;
int inside;
int i, j, k;
double xp, yp, x1, y1, x2, y2, x3, y3, xc, yc, r;
double xmin, xmax, ymin, ymax, xmid, ymid;
double dx, dy, dmax;
/* Allocate memory for the completeness list, flag for each triangle */
trimax = 4 * nv;
complete = new int[trimax];
/* Allocate memory for the edge list */
edges = new IEDGE[emax];
/*
Find the maximum and minimum vertex bounds.
This is to allow calculation of the bounding triangle
*/
xmin = pxyz[0].x;
ymin = pxyz[0].y;
xmax = xmin;
ymax = ymin;
for(i = 1; i < nv; i++){
if (pxyz[i].x < xmin) xmin = pxyz[i].x;
if (pxyz[i].x > xmax) xmax = pxyz[i].x;
if (pxyz[i].y < ymin) ymin = pxyz[i].y;
if (pxyz[i].y > ymax) ymax = pxyz[i].y;
}
dx = xmax - xmin;
dy = ymax - ymin;
dmax = (dx > dy) ? dx : dy;
xmid = (xmax + xmin) / 2.0;
ymid = (ymax + ymin) / 2.0;
/*
Set up the supertriangle
his is a triangle which encompasses all the sample points.
The supertriangle coordinates are added to the end of the
vertex list. The supertriangle is the first triangle in
the triangle list.
*/
pxyz[nv+0].x = xmid - 20 * dmax;
pxyz[nv+0].y = ymid - dmax;
pxyz[nv+1].x = xmid;
pxyz[nv+1].y = ymid + 20 * dmax;
pxyz[nv+2].x = xmid + 20 * dmax;
pxyz[nv+2].y = ymid - dmax;
v[0].p1 = nv;
v[0].p2 = nv+1;
v[0].p3 = nv+2;
complete[0] = false;
ntri = 1;
/*
Include each point one at a time into the existing mesh
*/
for(i = 0; i < nv; i++){
xp = pxyz[i].x;
yp = pxyz[i].y;
nedge = 0;
/*
Set up the edge buffer.
If the point (xp,yp) lies inside the circumcircle then the
three edges of that triangle are added to the edge buffer
and that triangle is removed.
*/
for(j = 0; j < ntri; j++){
if(complete[j])
continue;
x1 = pxyz[v[j].p1].x;
y1 = pxyz[v[j].p1].y;
x2 = pxyz[v[j].p2].x;
y2 = pxyz[v[j].p2].y;
x3 = pxyz[v[j].p3].x;
y3 = pxyz[v[j].p3].y;
inside = CircumCircle(xp, yp, x1, y1, x2, y2, x3, y3, xc, yc, r);
if (xc + r < xp)
// Suggested
// if (xc + r + EPSILON < xp)
complete[j] = true;
if(inside){
/* Check that we haven't exceeded the edge list size */
if(nedge + 3 >= emax){
emax += 100;
p_EdgeTemp = new IEDGE[emax];
for (int i = 0; i < nedge; i++) { // Fix by John Bowman
p_EdgeTemp[i] = edges[i];
}
delete []edges;
edges = p_EdgeTemp;
}
edges[nedge+0].p1 = v[j].p1;
edges[nedge+0].p2 = v[j].p2;
edges[nedge+1].p1 = v[j].p2;
edges[nedge+1].p2 = v[j].p3;
edges[nedge+2].p1 = v[j].p3;
edges[nedge+2].p2 = v[j].p1;
nedge += 3;
v[j] = v[ntri-1];
complete[j] = complete[ntri-1];
ntri--;
j--;
}
}
/*
Tag multiple edges
Note: if all triangles are specified anticlockwise then all
interior edges are opposite pointing in direction.
*/
for(j = 0; j < nedge - 1; j++){
for(k = j + 1; k < nedge; k++){
if((edges[j].p1 == edges[k].p2) && (edges[j].p2 == edges[k].p1)){
edges[j].p1 = -1;
edges[j].p2 = -1;
edges[k].p1 = -1;
edges[k].p2 = -1;
}
/* Shouldn't need the following, see note above */
if((edges[j].p1 == edges[k].p1) && (edges[j].p2 == edges[k].p2)){
edges[j].p1 = -1;
edges[j].p2 = -1;
edges[k].p1 = -1;
edges[k].p2 = -1;
}
}
}
/*
Form new triangles for the current point
Skipping over any tagged edges.
All edges are arranged in clockwise order.
*/
for(j = 0; j < nedge; j++) {
if(edges[j].p1 < 0 || edges[j].p2 < 0)
continue;
v[ntri].p1 = edges[j].p1;
v[ntri].p2 = edges[j].p2;
v[ntri].p3 = i;
complete[ntri] = false;
ntri++;
}
}
/*
Remove triangles with supertriangle vertices
These are triangles which have a vertex number greater than nv
*/
for(i = 0; i < ntri; i++) {
if(v[i].p1 >= nv || v[i].p2 >= nv || v[i].p3 >= nv) {
v[i] = v[ntri-1];
ntri--;
i--;
}
}
delete[] edges;
delete[] complete;
return 0;
}
int XYZCompare(const void *v1, const void *v2){
XYZ *p1, *p2;
p1 = (XYZ*)v1;
p2 = (XYZ*)v2;
if(p1->x < p2->x)
return(-1);
else if(p1->x > p2->x)
return(1);
else
return(0);
}
I didn't go with a debugger, but from the resulting triangles it seems that this is an accuracy/ambiguity problem.
When you are triangulating a square there are two ways to split it into triangles and both are OK from Delaunay criteria (circumscribed circle center is on border of triangle).
So if you evaluate every triangle independently you may sometimes get even 4 triangles (depending on implementation).
Normally in such cases I recommend to build algorithm as a series of questions which cannot produce contradicting answers. In this case the question is "to which point goes triangle based on edge (1,0)-(1,1)". But often this requires significant changes to the algorithm.
A quick fix usually involves adding some tolerances for comparisons and extra checks (like non-intersecting triangles). But usually it just makes problems rarer.
Most likely you didn't delete all the double edges, especially not the edges from same triangles but with vertices only in another order. The correct function is in the answer from #cMinor.
First question, I have tried to calculate the expression, di+1=di+2*Δy−2*Δx(yi+1−yi) for the four quadrants. Irrespective of the quadrant, the expression was found to be the same, including signs.
Am I right, or, there has been some mistakes in my calculations (hence, I am wrong)?
Second question, if this expression is only applicable for the first octet, how can I apply this to other octets? To me, there is no way to determine which octet I am working on. Coz, the value of m always represent two opposite octets. For example, if 0<m<1, it represents 1st and 5th octet. Right?
Thirdly, how can we determine the initial/starting value of di?
#include <iostream>
#include "utils.h"
void BresenhamLine(double x1, double y1, double x2, double y2, int color)
{
if(x1>x2 || y1>y2)
{
Swap(x1, x2);
Swap(y1, y2);
}
double x = x1;
double y = y1;
double dx = x2 - x1;
double dy = y2 - y1;
double dt = 2 * (dy - dx);
double ds = 2 * dy;
double d = 2*dy - dx;
PlotPixel(x, y, color);
if(dx>=dy)
{
while(x<=x2)
{
x++;
if(d<0)
{
d = d + ds;
}
else
{
y++;
d = d + dt;
}
PlotPixel(x, y, color);
}
}
else
{
while(y<=y2)
{
y++;
if(d<0)
{
x++;
d = d + dt;
}
else
{
d = d + ds;
}
PlotPixel(x, y, color);
}
}
}
int main()
{
int gm = DETECT;
int gd = DETECT;
initgraph(&gm, &gd, "");
double x1 = 0;
double y1 = 0;
double r = 50;
double x2 = 0;
double y2 = 0;
double signx = 0;
double signy = 0;
for(int theta=0 ; theta<=360 ; theta++)
{
x2 = r * cos(DegreeToRad((double) theta));
y2 = r * sin(DegreeToRad((double) theta));
x1 = 5 * cos(DegreeToRad((double) theta));
y1 = 5 * sin(DegreeToRad((double) theta));
BresenhamLine(x1, y1, x2, y2, YELLOW);
}
getch();
closegraph();
return 0;
}
The lines that go through 2nd and 4th quadrant are not showing up.
How to fix that with some minor changes in my code?
With this input: x1: 100 y1: -100 x2: -100 y2: 100
this logic:
if(x1>x2 || y1>y2)
{
Swap(x1, x2);
Swap(y1, y2);
}
fails.
This page is a good place to start. It shows code as well for 1 of the octants:
http://www.cs.helsinki.fi/group/goa/mallinnus/lines/bresenh.html
I think you need to swap the x if x1 > x2 and swap the y if y1 > y2 but not swap both if only 1 of those is true.
The external links section of the Wikipedia page contains several links to ready-made implementations that you can study.
Try this:
void BresenhamLine( double x1, double y1, double x2, double y2, int color )
{
const bool steep = (std::abs(y2 - y1) > std::abs(x2 - x1));
if(steep)
{
std::swap(x1, y1);
std::swap(x2, y2);
}
if(x1 > x2)
{
std::swap(x1, x2);
std::swap(y1, y2);
}
double dx = x2 - x1;
double dy = std::abs(y2 - y1);
double error = dx / 2;
int ystep = (y1 < y2) ? 1 : -1;
int y = (int)y1;
int maxX = (int)x2;
for(int x=(int)x1; x<maxX; x++)
{
if(steep)
{
PlotPixel(y, x, color);
}
else
{
PlotPixel(x, y, color);
}
error -= dy;
if(error < 0)
{
y += ystep;
error += dx;
}
}
}
I got this by slightly modifying the code here:http://rosettacode.org/wiki/Bitmap/Bresenham's_line_algorithm#C.2B.2B
What i want is that, I have info of 2 points, the starting x,y and mid point x,y and i need to find end line like until some kind of border, like window
here is what I do:
//function for calculating the end point from one location, to specific end location
//like a bullet moving forward in a line
//x,y start location(mouse), x2,y2(rect point location one of the 4) mid point, qx,qy end point(shadow or triangle draw location)
void screenEnd(int x, int y, int x2, int y2, int*qx,int*qy)
{
x = x2-x;
y = y2-y;
float tx = x2,ty = y2;
float result = atan2((float)y,(float)x) * 180 / PI;
float tempx = cos ( result * PI / 180.0 );
float tempy = sin ( result * PI / 180.0 );
bool check = true;
//this part needs optimization
while(check)
{
if(tx < 0|| ty < 0|| tx > 1280 || ty > 720)
{
check = false;
}
else
{
tx += tempx;
ty += tempy;
}
}
*qx = tx;
*qy = ty;
}
what I do is just increase point until it reaches the end.
Is there any way faster?
A classic window clipping task.
Consider a parametric equation where p is the point (x,y).
p(0) = x, y
p(0.5) = x2, y2
p(1) = x+2*(x2-x), y + 2*(y2-y)
p(t) = p(0) + t*(p(1) - p(0))
clip window = 0,0 to 720, 1280 (suspect you really want 719,1279)
The segment to draw initially ranges from t=0.0 to t=1.0. The segment is tested against each of the 4 sides of the bounding box, potentially reducing the t range. Maybe even eliminating all together.
Follows is some old code, enough to get you going.
#include <math.h>
int cliptest(int dz, int z, double *t0, double *t1) {
if (dz < 0) {
double t = ((double) z) / dz;
if (t > *t1)
return 0;
if (t > *t0)
*t0 = t;
} else if (dz > 0) {
double t = ((double) z) / dz;
if (t < *t0)
return 0;
if (t < *t1)
*t1 = t;
} else {
if (z < 0)
return 0;
}
return 1;
}
int clipper(int *px0, int *py0, int *px1, int *py1, int minx, int miny,
int maxx, int maxy) {
double t0, t1;
int dx, dy;
t0 = 0.0;
t1 = 1.0;
dy = *py1 - *py0;
dx = *px1 - *px0;
if (cliptest(-dx, *px0 - minx, &t0, &t1)
&& cliptest(dx, maxx - *px0, &t0, &t1)
&& cliptest(-dy, *py0 - miny, &t0, &t1)
&& cliptest(dy, maxy - *py0, &t0, &t1)) {
if (t1 < 1.0) {
*px1 = round(*px0 + t1*dx);
*py1 = round(*py0 + t1*dy);
}
if (t0 > 0.0) {
*px0 = round(*px0 + t0*dx);
*py0 = round(*py0 + t0*dy);
}
return 1;
}
return 0;
}
int x0 = x;
int y0 = y;
int x1 = x + 2*(x2-x); // Form end point
int y1 = x + 2*(y2-y);
if (clipper(&x0, &y0, &x1, &y1, 0, 0, 720, 1280))
Draw(x0, y0, x1, y2);
else
Handle_LineTotallyClippedOut();