我一直在阅读CUDA文档,看起来每个需要与OpenGL交互的缓冲区都需要在glBuffer中创建。
根据Nvidia编程指南,这必须按照以下方式完成:
GLuint positionsVBO;
struct cudaGraphicsResource* positionsVBO_CUDA;
int main() {
// Explicitly set device
cudaGLSetGLDevice(0);
// Initialize OpenGL and GLUT
...
glutDisplayFunc(display);
// Create buffer object and register it with CUDA
glGenBuffers(1, positionsVBO);
glBindBuffer(GL_ARRAY_BUFFER, &vbo);
unsigned int size = width * height * 4 * sizeof(float);
glBufferData(GL_ARRAY_BUFFER, size, 0, GL_DYNAMIC_DRAW);
glBindBuffer(GL_ARRAY_BUFFER, 0);
cudaGraphicsGLRegisterBuffer(&positionsVBO_CUDA, positionsVBO, cudaGraphicsMapFlagsWriteDiscard);
// Launch rendering loop
glutMainLoop();
}
void display() {
// Map buffer object for writing from CUDA
float4* positions;
cudaGraphicsMapResources(1, &positionsVBO_CUDA, 0);
size_t num_bytes;
cudaGraphicsResourceGetMappedPointer((void**)&positions, &num_bytes, positionsVBO_CUDA));
// Execute kernel
dim3 dimBlock(16, 16, 1);
dim3 dimGrid(width / dimBlock.x, height / dimBlock.y, 1);
createVertices<<<dimGrid, dimBlock>>>(positions, time, width, height);
// Unmap buffer object
cudaGraphicsUnmapResources(1, &positionsVBO_CUDA, 0);
// Render from buffer object
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glBindBuffer(GL_ARRAY_BUFFER, positionsVBO);
glVertexPointer(4, GL_FLOAT, 0, 0);
glEnableClientState(GL_VERTEX_ARRAY);
glDrawArrays(GL_POINTS, 0, width * height);
glDisableClientState(GL_VERTEX_ARRAY);
// Swap buffers
glutSwapBuffers();
glutPostRedisplay();
}
void deleteVBO() {
cudaGraphicsUnregisterResource(positionsVBO_CUDA);
glDeleteBuffers(1, &positionsVBO);
}
__global__ void createVertices(float4* positions, float time, unsigned int width, unsigned int height) {
// [....]
}
有没有办法将由cudaMalloc创建的内存空间直接提供给OpenGL?我已经编写了使用cuda的可工作代码,我想直接将我的float4数组放入OpenGL中。假设您已经有以下代码:
float4 *cd = (float4*) cudaMalloc(elements*sizeof(float4)).
do_something<<<16,1>>>(cd);
我想通过OpenGL显示do_something的输出。
顺便说一下:为什么每个时间步都要运行cudaGraphicsResourceGetMappedPointer函数?