如何在3D图形OpenGL中使用纹理

Question

正如汤米所说，我做了一些改变：

因此，我想说的是：

• 确保所有数组中至少包含11个元素；
• 启用纹理。

为什么此代码不会显示草的纹理？

首先，加载纹理图像：

QImage img;
if(!img.load("../g.jpg")){
    qDebug("Texture not found!");
}
else {
    t = QGLWidget::convertToGLFormat(img);
    qDebug("Textture Converted to GLFormat.");
}

然后调用draw方法，该方法应使用顶点数组绘制不平坦的地面：

 GLfloat vertices2[] = {
    2,2,0 , 2,-2,0 , -2,-2,0 , -2,2,0,

    0.7,0.6,0.6 , 0.4,0.9,0.1 , 0.1,0.2,0.3 , 0.8,0.3,0.8 ,

    0.5,-0.5,0.12 , 0,-0.9,0.4 , -0.8,-0.2,0.6
};

GLfloat colors[] = {35 , 73 , 12 , 10 , 6 , 41 , 32 , 10 , 13 , 35 , 42 , 86 , 12 , 2 , 6 ,9};

GLfloat TexCoords[] = {
    0.449,0.911,0.056,
    0.911,0.056,0.17,
    0.056,0.17,0.902,
    0.17,0.902,0.873,
    0.902,0.873,0.061,
    0.873,0.061,0.557,
    0.061,0.557,0.517,
    0.557,0.517,0.131,
    0.517,0.131,0.855,
    0.131,0.855,0.533,
    0.855,0.533,0.896,
    0.533,0.896,0.178,
    0.896,0.178,0.303,
    0.178,0.303,0.358,
    0.303,0.358,0.064,
    0.358,0.064,0.156,
    0.064,0.156,0.199,
    0.156,0.199,0.32,
    0.199,0.32,0.636,
    0.32,0.636,0.306,
    0.636,0.306,0.445,
    0.306,0.445,0.166,
    0.445,0.166,0.572,
    0.166,0.572,0.249,
    0.572,0.249,0.29,
    0.249,0.29,0.389,
    0.29,0.389,0.79,
    0.389,0.79,0.567,
    0.79,0.567,0.692,
    0.567,0.692,0.202,
    0.692,0.202,0.913,
    0.202,0.913,0.498
};

GLfloat NormalP[] = {     0.383,0.29,0.746,
                          0.29,0.746,0.301,
                          0.746,0.301,0.675,
                          0.301,0.675,0.541,
                          0.675,0.541,0.677,
                          0.541,0.677,0.896,
                          0.677,0.896,0.915,
                          0.896,0.915,0.295,
                          0.915,0.295,0.423,
                          0.295,0.423,0.76,
                          0.423,0.76,0.89,
                    };

GLubyte indices[]  = {
    1,2,9 , 9,8,1 , 1,8,7 , 7,1,0 ,
    0,7,4 , 0,4,5 , 0,5,3 , 3,5,6 ,
    3,6,10 , 3,10,2 , 2,10,9 , 9,6,10 ,
    6,5,4 , 6,4,7 , 6,7,8 , 6,8,9
};

GLuint mHandle;
glGenTextures(1, &mHandle);
glBindTexture(GL_TEXTURE_2D, mHandle);

glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_R, GL_CLAMP_TO_EDGE);
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_S, GL_CLAMP_TO_EDGE);
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_T, GL_CLAMP_TO_EDGE);
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MIN_FILTER, GL_LINEAR_MIPMAP_LINEAR);
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MAG_FILTER, GL_LINEAR);

glTexImage2D(GL_TEXTURE_2D,0,GL_RGBA,t.width(),t.height(),0,GL_RGBA,GL_UNSIGNED_BYTE,t.bits());

glEnableClientState(GL_VERTEX_ARRAY);
glEnableClientState(GL_COLOR_ARRAY);
glEnableClientState(GL_TEXTURE_COORD_ARRAY);
glEnableClientState(GL_NORMAL_ARRAY);

glColorPointer(3,GL_FLOAT,0,colors);
glVertexPointer(3, GL_FLOAT, 0, vertices2);
glTexCoordPointer(3,GL_FLOAT,0,TexCoords);
glNormalPointer(GL_FLOAT,0,NormalP);

glDrawElements(GL_TRIANGLES,54, GL_UNSIGNED_BYTE, indices);
glDisableClientState(GL_VERTEX_ARRAY);

Answer 1

在固定管道中，仅在启用纹理后才应用纹理化。 您不会显示对glEnable(GL_TEXTURE_2D)的调用。

话虽这么说，但看起来您的阵列大小可能不正确。

glTexCoordPointer(3,GL_FLOAT,0,TexCoords); ... glDrawElements(GL_TRIANGLES,54, GL_UNSIGNED_BYTE, indices);的净效果glTexCoordPointer(3,GL_FLOAT,0,TexCoords); ... glDrawElements(GL_TRIANGLES,54, GL_UNSIGNED_BYTE, indices); glTexCoordPointer(3,GL_FLOAT,0,TexCoords); ... glDrawElements(GL_TRIANGLES,54, GL_UNSIGNED_BYTE, indices); 是11套纹理坐标将被读取TexCoords因为在最大的指数indices为10 。 但是，假设您在3d中指定纹理坐标数组，则其中只有两个纹理坐标。

因此，我想说的是：

• 确保所有数组中至少包含11个元素；
• 启用纹理。

如果仅使用2d纹理并且对纹理矩阵堆栈不做任何聪明的事情，您可能还想切换到2d提供纹理坐标。

按照jp的说法，这是为什么现在被标记为社区Wiki的公然语：创建新纹理ID后，您也无法绑定它。 OpenGL有点奇怪，因为在显示列表被证明没有最初希望的属性后，纹理ID便被固定住了，因此假设您只有一个纹理，那么即使您也从未绑定过纹理，您就可以不做任何事情而无需生成纹理。 但是，如果您对此有所了解并生成它们，请确保您绑定它们。 生成不会隐式绑定。

另一个问题是您启用了mipmapping作为采样参数的一部分：

glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MIN_FILTER, GL_LINEAR_MIPMAP_LINEAR);

但纹理没有mipmap。 要生成mipmap，请在glTexImage2D()之后添加以下调用：

glGenerateMipmap(GL_TEXTURE_2D);