Describing the Task

In this series of articles, I want to show the way OpenGL works by writing its clone (a much simplified one). Surprisingly enough, I often meet people who cannot overcome the initial hurdle of learning OpenGL / DirectX. Thus, I have prepared a short series of six lectures, after which my students show quite good renderers.

So, the task is formulated as follows: Using no third-party libraries (especially graphic ones), get something like this picture:

Warning you, that it is training material that will generally repeat the structure of the OpenGL library. It will be a software renderer. I do not want to show how to write applications for OpenGL. I want to show how OpenGL is built. I am deeply convinced that it is impossible to write efficient applications using 3D libraries without understanding this.

I will try to make the final code not longer than 500 lines. My students need 10 to 20 programming hours to begin making such renderers. At the input, we get a test file with a polygonal wire + pictures with textures. At the output, we’ll get a rendered model. No graphical interface, the program simply generates an image.

Since the goal is to minimize external dependencies, I give my students just one class that allows working with TGA files. It’s one of the simplest formats that supports images in RGB/RGBA/black and white formats. So, as a starting point, we’ll obtain a simple way to work with pictures. You should note that the only functionality available at the very beginning (in addition to loading and saving images) is the capability to set the color of one pixel.

There are no functions for drawing line segments and triangles. We’ll have to do all of this by hand.

I provide my source code that I write in parallel with students. But I would not recommend using it, as this doesn’t make sense.

The entire code is available on github, and here you will find the source code I give to my students.

#include "tgaimage.h"
const TGAColor white = TGAColor(255, 255, 255, 255);
const TGAColor red   = TGAColor(255, 0,   0,   255);
int main(int argc, char** argv) {
        TGAImage image(100, 100, TGAImage::RGB);
        image.set(52, 41, red);
        image.flip_vertically(); // i want to have the origin at the left bottom corner of the image
        image.write_tga_file("output.tga");
        return 0;
}

output.tga should look something like this:

Bresenham’s Line Algorithm

The goal of the first lecture is to render the wire mesh. To do this, we should learn how to draw line segments.

We can simply read what Bresenham’s line algorithm is, but let’s write code ourselves.

How does the simplest code that draws a line segment between (x0, y0) and (x1, y1) points look like?

Apparently, something like this

void line(int x0, int y0, int x1, int y1, TGAImage &image, TGAColor color) {
    for (float t=0.; t<1.; t+=.01) {
        int x = x0*(1.-t) + x1*t;
        int y = y0*(1.-t) + y1*t;
        image.set(x, y, color);
    }
}

The snapshot of the code is available on github.


The problem with this code (in addition to efficiency) is the choice of the constant, which I took equal to .01.

If we take it equal to .1, our line segment will look like this:

We can easily find the necessary step: it’s just the number of pixels to be drawn.

The simplest (with errors!) code looks something like the following:

void line(int x0, int y0, int x1, int y1, TGAImage &image, TGAColor color) {
    for (int x=x0; x<=x1; x++) {
        float t = (x-x0)/(float)(x1-x0);
        int y = y0*(1.-t) + y1*t;
        image.set(x, y, color);
    }
}

Caution! The first source of errors in such code of my students is the integer division, like (x-x0)/(x1-x0).

Then, if we try to draw the following lines with this code:

line(13, 20, 80, 40, image, white);
line(20, 13, 40, 80, image, red);
line(80, 40, 13, 20, image, red);

It turns out that one line is good, the second one is with holes, and there’s no third line at all. Note that the first and the second lines (in the code) give the same line of different colors. We have already seen the white one, it is drawn well. I was hoping to change the color of the white line to red, but could not do it. It’s a test for symmetry: the result of drawing a line segment should not depend on the order of points: the (a,b) line segment should be exactly the same as the (b,a) line segment.


There are holes in one of the line segments due to the fact that its height is greater than the width.

My students often suggest the following fix: if (dx>dy) {for (int x)} else {for (int y)}.

Holy cow!

void line(int x0, int y0, int x1, int y1, TGAImage &image, TGAColor color) {
    bool steep = false;
    if (std::abs(x0-x1)x1) { // make it left-to-right
        std::swap(x0, x1);
        std::swap(y0, y1);
    }
    for (int x=x0; x<=x1; x++) {
        float t = (x-x0)/(float)(x1-x0);
        int y = y0*(1.-t) + y1*t;
        if (steep) {
            image.set(y, x, color); // if transposed, de-transpose
        } else {
            image.set(x, y, color);
        }
    }
}


This code works great. That’s exactly the kind of complexity I want to see in the final version or our renderer.
It is definitely inefficient (multiple divisions, and the like), but it is short and readable.
Note that it has no asserts and no checks on going beyond the borders, which is bad.

But I try not to overload this particular code, as it is read a lot. At the same time, I systematically remind of the necessity to perform checks.

So, the previous code works fine, but we can optimize it.

Optimization is a dangerous thing. We should be clear about the platform the code will run on.
Optimize the code for a graphics card or just for a CPU — are completely different things.
Before and during any optimization, the code should be profiled.
Try to guess, which operation is the most recourse-intensive operation here?

For tests, 1,000,000 times I draw 3 line segments we have drawn before. My CPU is Intel® Core(TM) i5-3450 CPU @ 3.10GHz.

For each pixel, this code calls the TGAColor copy constructor.

Which is 1000000 * 3 line segments * approximately 50 pixels per line segment. Quite a lot of calls, isn’t it?

Where to start with optimization?

The profiler will tell us.

I compiled the code with g++ -ggdb -g3 -pg -O0 keys, and then ran gprof:

%   cumulative   self              self     total
 time   seconds   seconds    calls  ms/call  ms/call  name
 69.16      2.95     2.95  3000000     0.00     0.00  line(int, int, int, int, TGAImage&, TGAColor)
 19.46      3.78     0.83 204000000     0.00     0.00  TGAImage::set(int, int, TGAColor)
  8.91      4.16     0.38 207000000     0.00     0.00  TGAColor::TGAColor(TGAColor const&)
  1.64      4.23     0.07        2    35.04    35.04  TGAColor::TGAColor(unsigned char, unsigned char, unsigned char, unsigned char)
  0.94      4.27     0.04                             TGAImage::get(int, int)

10% of the time are spent on copying the color.

But 70% are performed in calling line()! That’s where we will optimize.


We should note that each division has the same divisor. Let’s take it out of the loop.
The error variable gives is the distance to the best straight line from our current (x, y) pixel.
Each time error is greater than one pixel, we increase (or decrease) y by one, and decrease the error by one as well.

The code is available here.

void line(int x0, int y0, int x1, int y1, TGAImage &image, TGAColor color) {
    bool steep = false;
    if (std::abs(x0-x1)x1) {
        std::swap(x0, x1);
        std::swap(y0, y1);
    }
    int dx = x1-x0;
    int dy = y1-y0;
    float derror = std::abs(dy/float(dx));
    float error = 0;
    int y = y0;
    for (int x=x0; x<=x1; x++) {
        if (steep) {
            image.set(y, x, color);
        } else {
            image.set(x, y, color);
        }
        error += derror;
        if (error>.5) {
            y += (y1>y0?1:-1);
            error -= 1.;
        }
    }
}
%   cumulative   self              self     total
 time   seconds   seconds    calls  ms/call  ms/call  name
 38.79      0.93     0.93  3000000     0.00     0.00  line(int, int, int, int, TGAImage&, TGAColor)
 37.54      1.83     0.90 204000000     0.00     0.00  TGAImage::set(int, int, TGAColor)
 19.60      2.30     0.47 204000000     0.00     0.00  TGAColor::TGAColor(int, int)
  2.09      2.35     0.05        2    25.03    25.03  TGAColor::TGAColor(unsigned char, unsigned char, unsigned char, unsigned char)
  1.25      2.38     0.03                             TGAImage::get(int, int)

Why do we need floating points? The only reason is one division by dx and comparison with .5 in the loop body.

We can get rid of the floating point by replacing the error variable with another one. Let’s call it error2, and assume it is equal to error*dx*2.

Here’s the equivalent code:

void line(int x0, int y0, int x1, int y1, TGAImage &image, TGAColor color) {
    bool steep = false;
    if (std::abs(x0-x1)x1) {
        std::swap(x0, x1);
        std::swap(y0, y1);
    }
    int dx = x1-x0;
    int dy = y1-y0;
    int derror2 = std::abs(dy)*2;
    int error2 = 0;
    int y = y0;
    for (int x=x0; x<=x1; x++) {
        if (steep) {
            image.set(y, x, color);
        } else {
            image.set(x, y, color);
        }
        error2 += derror2;
        if (error2 > dx) {
            y += (y1>y0?1:-1);
            error2 -= dx*2;
        }
    }
}
%   cumulative   self              self     total
 time   seconds   seconds    calls  ms/call  ms/call  name
 42.77      0.91     0.91 204000000     0.00     0.00  TGAImage::set(int, int, TGAColor)
 30.08      1.55     0.64  3000000     0.00     0.00  line(int, int, int, int, TGAImage&, TGAColor)
 21.62      2.01     0.46 204000000     0.00     0.00  TGAColor::TGAColor(int, int)
  1.88      2.05     0.04        2    20.02    20.02  TGAColor::TGAColor(unsigned char, unsigned char, unsigned char, unsigned char)

Now, it’s enough to remove unnecessary copies during the function call by passing the color by reference (or just enable the compilation flag -O3), and it’s done. Not a single multiplication or division in code. The execution time has decreased from 2.95 to 0.64 seconds.

Wire Render

So now we are ready to create a wire render. You can find the snapshot of the code and the test model here.

I used the wavefront obj format of the file to store model. All we need for the render is read from the file the array of vertices of the following type:

v 0.608654 -0.568839 -0.416318
[...]
are x,y,z coordinates, one vertex per file line

and faces

f 1193/1240/1193 1180/1227/1180 1179/1226/1179
[...]

We are interested in the first number after each space. It is the number of the vertex in the array that we have read before. Thus, this line says that 1193, 1180 and 1179 vertices form a triangle.

The model.cpp file contains a simple parser.

Write the following loop to our main.cpp and voila, our wire renderer is ready.

for (int i=0; infaces(); i++) {
        std::vector face = model->face(i);
        for (int j=0; j<3; j++) {
            Vec3f v0 = model->vert(face[j]);
            Vec3f v1 = model->vert(face[(j+1)%3]);
            int x0 = (v0.x+1.)*width/2.;
            int y0 = (v0.y+1.)*height/2.;
            int x1 = (v1.x+1.)*width/2.;
            int y1 = (v1.y+1.)*height/2.;
            line(x0, y0, x1, y1, image, white);
        }
    }

Next time we will draw 2D triangles and improve our renderer.

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