Rendering visually perfect squares in OpenGL? - opengl

Question

In OpenGL's fixed pipeline, by default, specifying vertex coordinates using glVertex3f is equivalent to specifying a location between -1.0 and +1.0 in screen space. Therefore, given a set of 4 perfectly adjacent screen-space vertices using GL_TRIANGLE_STRIP (or even GL_QUADS), and unless your window is already perfectly square, you will always render a rectangle instead of a perfect square...
Knowing the width, height and aspect ratio of a window, is there some way to correct this?
I have tried multiplying the vertex coordinates by the aspect ratio, which unfortunately seemed to achieve the same visual effect.
Here's the full source code I'm currently using:
#include "main.h"
#pragma comment(lib, "glut32.lib")
int g_width = 800;
int g_height = 600;
int g_aspectRatio = double(g_width) / double(g_height);
bool g_bInitialized = false;
int main(int argc, char **argv)
{
glutInit(&argc, argv);
glutInitDisplayMode(GLUT_DEPTH | GLUT_DOUBLE | GLUT_RGBA);
glutInitWindowPosition(0, 0);
glutInitWindowSize(g_width, g_height);
glutCreateWindow("OpenGL Test App");
glutDisplayFunc(onRender);
glutReshapeFunc(onSize);
glutIdleFunc(onRender);
glutMainLoop();
return 0;
}
void onInit()
{
glFrontFace(GL_CW);
}
void onRender()
{
if(!g_bInitialized)
onInit();
static float angle = 0.0f;
const float p = 0.5f * g_aspectRatio;
glLoadIdentity();
gluLookAt(
0.0f, 0.0f, 10.0f,
0.0f, 0.0f, 0.0f,
0.0f, 1.0f, 0.0f
);
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glScalef(1, -1, 1); // Flip the Y-axis
glRotatef(angle, 0.0f, 1.0f, 0.0f);
glBegin(GL_TRIANGLE_STRIP);
{
glColor4f(1.0, 0.0, 0.0, 1.0); // Red
glVertex3f(-p, -p, 0.0); // Top-Left
glColor4f(0.0, 1.0, 0.0, 1.0); // Green
glVertex3f(p, -p, 0.0); // Top-Right
glColor4f(0.0, 0.0, 1.0, 1.0); // Blue
glVertex3f(-p, p, 0.0); // Bottom-Left
glColor4f(1.0, 1.0, 0.0, 1.0); // Yellow
glVertex3f(p, p, 0.0); // Bottom-Left
}
glEnd();
angle += 0.6f;
glutSwapBuffers();
}
void onSize(int w, int h)
{
g_width = max(w, 1);
g_height = max(h, 1);
g_aspectRatio = double(g_width) / double(g_height);
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
glViewport(0, 0, w, h);
gluPerspective(45, g_aspectRatio, 1, 1000);
glMatrixMode(GL_MODELVIEW);
}
EDIT:
This has been solved... In the above code, I had defined g_aspectRatio as an int instead of a floating-point value. Therefore, it's value was always 1...

Answer 1

In my (old) experience, that's just why you have an aspect ratio argument to gluPerspective().
The manual page says:
In general, the aspect ratio in gluPerspective should match
the aspect ratio of the associated viewport. For example, aspect = 2.0
means the viewer's angle of view is twice as wide in x as it is in y.
If the viewport is twice as wide as it is tall, it displays the image
without distortion.
Check your g_aspectRatio value.

Answer 2

by default, specifying vertex coordinates using glVertex3f is equivalent to specifying a location between -1.0 and +1.0 in screen space
Wrong. Coordinates passed to OpenGL through glVertex or a glVertexPointer vertex array are in model space. The transformation to screen space happens by transforming into view space by the modelview matrix and from view space to clip space by the projection matrix. Then clipping is applied and the perspective divide applied to reach normalized coordinate space.
Hence the value range for glVertex can be whatever you like it to be. By applying the right projection matrix you get your view space to be in [-aspect; aspect]×[-1, 1] if you like that.
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
glOrtho(-aspect, aspect, -1, 1, -1, 1);