绘制一个简单的2D矩形网格

时间:2014-08-14 22:01:32

标签: c++ opengl 2d

基本上我想实现的是一个紧挨着的矩形集合,我可以从一个数组中控制。我似乎无法找到关于这个愚蠢简单的事情的任何可靠信息,因为大多数教程都需要我太大的帮助才能找到帮助。到目前为止,我甚至无法使它渲染网格,它使它非常大。 (出于某种原因?)

来源

#include <cstdlib>
#include <iostream>
#include <GLUT/GLUT.h>

void display();
void resize(int, int);
void timer(int);

#define SCREEN 512
#define REFRESH 30

const int GRID = 8;

/* Main function: GLUT runs as a console application starting at main()  */
int main(int argc, char** argv) {
    glutInit(&argc, argv);                 // Initialize GLUT
    glutCreateWindow("OpenGL Setup Test"); // Create a window with the given title
    glutInitWindowSize(SCREEN, SCREEN);   // Set the window's initial width & height
    glutTimerFunc(0, timer, 0);
    glutInitWindowPosition(0, SCREEN); // Position the window's initial top-left corner
    glutDisplayFunc(display); // Register display callback handler for window re-paint
    glutInitDisplayMode(GLUT_DOUBLE);
    glutReshapeFunc(resize);
    glutMainLoop();           // Enter the event-processing loop
    return 0;
}

/* Handler for window-repaint event. Call back when the window first appears and
 whenever the window needs to be re-painted. */
void display() {
    glClearColor(0.0f, 0.0f, 0.0f, 1.0f); // Set background color to black and opaque
    glClear(GL_COLOR_BUFFER_BIT);         // Clear the color buffer (background)
    glMatrixMode(GL_MODELVIEW);     // To operate on Model-View matrix
    glLoadIdentity();

    for(int x=0; x < SCREEN; x+=GRID)
    {
        for(int y=0; y < SCREEN; y+=GRID)
        {
            // Draw a Red 1x1 Square centered at origin
            glPushMatrix();
            glTranslatef((1.0f/x)*GRID, (1.0f/y)*GRID, 0);

            //std::cout << (GLfloat)1/x << " : " << (GLfloat)1/y << std::endl;

            glBegin(GL_QUADS);              // Each set of 4 vertices form a quad
            glColor3f(1.0f, 0.0f, 0.0f); // Red
            glVertex2f(-1.0f/GRID, -1.0f/GRID);    // x, y
            glVertex2f( 1.0f/GRID, -1.0f/GRID);
            glVertex2f( 1.0f/GRID,  1.0f/GRID);
            glVertex2f(-1.0f/GRID,  1.0f/GRID);
            glEnd();

            glPopMatrix();
        }
    }

    /*glBegin(GL_QUADS);              // Each set of 4 vertices form a quad
    glColor3f(1.0f, 0.0f, 0.0f); // Red
    glVertex2f(-0.005, -0.005);    // x, y
    glVertex2f( 0.005, -0.005);
    glVertex2f( 0.005,  0.005);
    glVertex2f(-0.005,  0.005);
    glEnd();*/

    glFlush();  // Render now
}

/* Called back when timer expired */
void timer(int value) {
    glutPostRedisplay();      // Post re-paint request to activate display()
    glutTimerFunc(REFRESH, timer, 0); // next Timer call milliseconds later
}

void resize(int width, int height) {
    // we ignore the params and do:
    glutReshapeWindow( 600, 600);
    gluOrtho2D(0, 0, width, height);
    glTranslatef(-0.5, -0.5, 0);
    glViewport(0, 0, width, height);
    glMatrixMode(GL_PROJECTION);
    glLoadIdentity();
}

它看起来像什么

What I'm seeing!

1 个答案:

答案 0 :(得分:6)

让我们首先定义我认为你的价值观意味着什么,因为你好像没有按照你的意图使用它们:

  • SCREEN:窗口大小(以像素为单位)。
  • GRID:每个四边形的大小,以像素为单位。

因此,每个方向适合的四边形数量为SCREEN / GRID。在绘制循环之前,可以计算一次该值:

int quadCount = SCREEN / GRID;

下一个重要的部分是理解OpenGL坐标系。如果您不应用任何转换(并且实际上不需要用于您的用例),则映射到窗口的xy的范围来自{ {1}}到-1.0f。因此,要填充窗口,所有四边形的坐标都需要在两个方向上填充1.0f范围。

由于您需要生成的坐标范围为[-1.0f, 1.0f](从2.0f-1.0f),因此此坐标系中每个四边形的宽度/高度为1.0f 。我会在嵌套循环之外计算一次这个值:

2.0f / quadCount

现在,要将四边形放置在给定位置,它们的大小需要乘以它们的索引。再次记住,坐标从float quadSize = 2.0f / static_cast<float>(quadCount); 开始。因此渲染代码可能如下所示:

(-1.0f, -1.0f)