Opengl pixel perfect 2D drawing

Question

I'm working on a 2d engine. It already works quite good, but I keep getting pixel-errors.

For example, my window is 960x540 pixels, I draw a line from (0, 0) to (959, 0). I would expect that every pixel on scan-line 0 will be set to a color, but no: the right-most pixel is not drawn. Same problem when I draw vertically to pixel 539. I really need to draw to (960, 0) or (0, 540) to have it drawn.

As I was born in the pixel-era, I am convinced that this is not the correct result. When my screen was 320x200 pixels big, I could draw from 0 to 319 and from 0 to 199, and my screen would be full. Now I end up with a screen with a right/bottom pixel not drawn.

This can be due to different things: where I expect the opengl line primitive is drawn from a pixel to a pixel inclusive, that last pixel just is actually exclusive? Is that it? my projection matrix is incorrect? I am under a false assumption that when I have a backbuffer of 960x540, that is actually has one pixel more? Something else?

Can someone please help me? I have been looking into this problem for a long time now, and every time when I thought it was ok, I saw after a while that it actually wasn't.

Here is some of my code, I tried to strip it down as much as possible. When I call my line-function, every coordinate is added with 0.375, 0.375 to make it correct on both ATI and nvidia adapters.

int width = resX();
int height = resY();

for (int i = 0; i < height; i += 2)
    rm->line(0, i, width - 1, i, vec4f(1, 0, 0, 1));
for (int i = 1; i < height; i += 2)
    rm->line(0, i, width - 1, i, vec4f(0, 1, 0, 1));

// when I do this, one pixel to the right remains undrawn

void rendermachine::line(int x1, int y1, int x2, int y2, const vec4f &color)
{
    ... some code to decide what std::vector the coordinates should be pushed into
    // m_z is a z-coordinate, I use z-buffering to preserve correct drawing orders
    // vec2f(0, 0) is a texture-coordinate, the line is drawn without texturing
    target->push_back(vertex(vec3f((float)x1 + 0.375f, (float)y1 + 0.375f, m_z), color, vec2f(0, 0)));
    target->push_back(vertex(vec3f((float)x2 + 0.375f, (float)y2 + 0.375f, m_z), color, vec2f(0, 0)));
}

void rendermachine::update(...)
{
    ... render target object is queried for width and height, in my test it is just the back buffer so the window client resolution is returned
    mat4f mP;
    mP.setOrthographic(0, (float)width, (float)height, 0, 0, 8000000);

    ... all vertices are copied to video memory

    ... drawing
    if (there are lines to draw)
        glDrawArrays(GL_LINES, (int)offset, (int)lines.size());

    ...
}

// And the (very simple) shader to draw these lines

// Vertex shader
    #version 120
    attribute vec3 aVertexPosition;
    attribute vec4 aVertexColor;
    uniform mat4 mP;
    varying vec4 vColor;
    void main(void) {
        gl_Position = mP * vec4(aVertexPosition, 1.0);
        vColor = aVertexColor;
    }

// Fragment shader
    #version 120
    #ifdef GL_ES
    precision highp float;
    #endif
    varying vec4 vColor;
    void main(void) {
        gl_FragColor = vColor.rgb;
    }

I'm beginning to think that it's the "point exclusive" thing I'm talking about. If you use QT or C++ Builder/Delphi or even windows API to draw lines, the pixels are drawn excluding the end-point. So it's possible that the same is done in opengl, and probably also when drawing triangles? When you draw a rectangle with QT, builder, ..., you also get the bottom right corner excluded... — scippie
The following test shows agreement with what I said above: when I use GL_POINT's to fill the pixels at all the corners of the window, these coordinates are as I expect them to be: (0,0), (width-1,height-1). This shows me that my way of drawing is correct and that line (and possibly triangle) will just not include the end-point. Can someone concur? — scippie
related stackoverflow.com/questions/5467218/opengl-2d-hud-over-3d — Ciro Santilli 新疆再教育营六四事件法轮功郝海东
related but more generic: stackoverflow.com/questions/7922526/… — Ciro Santilli 新疆再教育营六四事件法轮功郝海东

lurker number 5 lurker number 5 · Accepted Answer · 2012-04-06T09:29:50

In OpenGL, lines are rasterized using the "Diamond Exit" rule. This is almost the same as saying that the end coordinate is exclusive, but not quite...

This is what the OpenGL spec has to say: http://www.opengl.org/documentation/specs/version1.1/glspec1.1/node47.html

Also have a look at the OpenGL FAQ, http://www.opengl.org/archives/resources/faq/technical/rasterization.htm, item "14.090 How do I obtain exact pixelization of lines?". It says "The OpenGL specification allows for a wide range of line rendering hardware, so exact pixelization may not be possible at all."

Many will argue that you should not use lines in OpenGL at all. Their behaviour is based on how ancient SGI hardware worked, not on what makes sense. (And lines with widths >1 are nearly impossible to use in a way that looks good!)

Opengl pixel perfect 2D drawing

2 Answers