Question

正如汤米所说，我做了一些改变：

所以，我说步骤将是：


确保所有数组中至少包含11个元素;

启用纹理。

为什么这段代码不会显示草纹？

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，仅在启用时才应用纹理。您没有显示对glEnable(GL_TEXTURE_2D)的调用。

话虽如此，看起来您的阵列可能也不正确。

glTexCoordPointer(3,GL_FLOAT,0,TexCoords); ... glDrawElements(GL_TRIANGLES,54, GL_UNSIGNED_BYTE, indices);的净效果是将从TexCoords读取11组纹理坐标，因为indices中的最大索引是10。但是，您的纹理坐标数组中只有两个纹理坐标，前提是您在3d中指定它们。

所以，我会说步骤是：

确保所有数组中至少包含11个元素;
启用纹理。

如果您只使用2d纹理并且没有使用纹理矩阵堆栈做任何巧妙的事情，您可能还希望在2d中切换到提供纹理坐标。

根据j-p，明显的原因是为什么现在将其标记为社区维基：您在创建它之后也未能绑定新的纹理ID。 OpenGL有点奇怪，因为纹理ID在显示列表后被证明没有最初希望的属性，所以你可以在没有生成纹理的情况下完成所有事情，前提是你也永远不会绑定一个纹理，假设你只有一个纹理。但是如果你对它有所了解并生成它们，那么请确保你绑定它们。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);

如何在3d图形OpenGL中使用纹理

1 个答案: