我正在尝试将PBO复制到启用自动映射的纹理中,但似乎只生成顶级纹理(换句话说,没有发生mipmapping)。
我正在用
//Generate a buffer ID called a PBO (Pixel Buffer Object)
glGenBuffers(1, pbo);
//Make this the current UNPACK buffer
glBindBuffer(GL_PIXEL_UNPACK_BUFFER, *pbo);
//Allocate data for the buffer. 4-channel 8-bit image
glBufferData(GL_PIXEL_UNPACK_BUFFER, size_tex_data, NULL, GL_DYNAMIC_COPY);
cudaGLRegisterBufferObject(*pbo);
我正在用
// Enable Texturing
glEnable(GL_TEXTURE_2D);
// Generate a texture identifier
glGenTextures(1,textureID);
// Make this the current texture (remember that GL is state-based)
glBindTexture( GL_TEXTURE_2D, *textureID);
// Allocate the texture memory. The last parameter is NULL since we only
// want to allocate memory, not initialize it
glTexImage2D(GL_TEXTURE_2D, 0, GL_RGBA_FLOAT32_ATI, size_x, size_y, 0,
GL_RGBA, GL_FLOAT, NULL);
// Must set the filter mode, GL_LINEAR enables interpolation when scaling
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_S, GL_REPEAT);
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_T, GL_REPEAT);
glTexParameteri(GL_TEXTURE_2D,GL_TEXTURE_MIN_FILTER,GL_LINEAR_MIPMAP_LINEAR);
glTexParameteri(GL_TEXTURE_2D,GL_TEXTURE_MAG_FILTER,GL_LINEAR);
glTexParameteri(GL_TEXTURE_2D, GL_GENERATE_MIPMAP, GL_TRUE);
稍后,在内核中,我修改pbo时使用了如下方法:
float4* aryPtr = NULL;
cudaGLMapBufferObject((void**)&aryPtr, *pbo);
//Pixel* gpuPixelsRawPtr = thrust::raw_pointer_cast(&gpuPixels[0]);
//... do some cuda stuff to aryPtr ...
//If we don't unmap the PBO then OpenGL won't be able to use it:
cudaGLUnmapBufferObject(*pbo);
现在,在我使用上面生成的纹理绘制到屏幕之前,我调用:
(请注意,rtmipmaptex=*上面的textureid和rtresultpbo=*上面的pbo)
glEnable(GL_DEPTH);
glEnable(GL_TEXTURE_2D);
glBindTexture(GL_TEXTURE_2D, rtMipmapTex);
glBindBuffer(GL_PIXEL_UNPACK_BUFFER, rtResultPBO);
glTexSubImage2D(GL_TEXTURE_2D, 0, 0, 0, canvasSize.x, canvasSize.y, GL_RGBA, GL_FLOAT, NULL);
这一切工作良好,正确显示纹理。但是,如果我把最后一行改成
glTexSubImage2D(GL_TEXTURE_2D, 1, 0, 0, canvasSize.x, canvasSize.y, GL_RGBA, GL_FLOAT, NULL);
据我所知,
应该
给我看第一层而不是纹理金字塔中的第零层纹理,我只得到一个空白的白色纹理。
如何从PBO复制纹理以触发自动mipmapping?
谢谢。
