c++ - OpenGL ES 绘图形状

Question

我正在学习 OpenGL ES 以呈现原生 android 开发。我可以画一个三角形。但是我不能画出一个以上的三角形。我只想绘制一个矩形，但如果我告诉 glVertexPointer 函数超过 3 个顶点，那么它不会绘制。我尝试使用 GL_TRIANGLES 和 GL_TRIANGLE_STRIP。

我做错了什么？

struct Quad
{
    GLfloat         Vertices[18];
    const GLbyte    nNumVerts;

    Quad(GLfloat i_fWidth, GLfloat i_fHeight) : nNumVerts(6)
    {
        GLfloat wide = i_fWidth / 2;
        GLfloat high = i_fHeight / 2;

        Vertices[0] = -wide;    Vertices[1] = high;     Vertices[2] = 0.0f;
        Vertices[3] = -wide;    Vertices[4] = -high;    Vertices[5] = 0.0f;
        Vertices[6] = wide;     Vertices[7] = -high;    Vertices[8] = 0.0f;

        Vertices[9] = -wide;    Vertices[10] = high;    Vertices[11] = 0.0f;
        Vertices[12] = wide;    Vertices[13] = -high;   Vertices[14] = 0.0f;
        Vertices[15] = wide;    Vertices[16] = high;    Vertices[17] = 0.0f;
    }
};

void Renderer::Render()
{
    glClear(GL_COLOR_BUFFER_BIT);

    glEnableClientState(GL_VERTEX_ARRAY);

    // If I change "m_Quad.nNumVerts" to '3' instead of '6' it will draw a triangle
    // Anything higher than '3' and it doesn't draw anything
    glVertexPointer(m_Quad.nNumVerts, GL_FLOAT, 0, m_Quad.Vertices);

    glDrawArrays(GL_TRIANGLES, 0, m_Quad.nNumVerts);

    eglSwapBuffers(m_pEngine->pDisplay, m_pEngine->pSurface);
}

Answer 1

glVertexPointer 的第一个参数是每个顶点的值的数量，在这种情况下它应该保持在 3。

glVertexPointer(3, GL_FLOAT, 0, m_Quad.Vertices);