我是CUDA编程的新手，我需要帮助编写一个程序来将图像存储在内存缓冲区中。我尝试修改CUDA-OpenGL互操作示例中的代码，在CUDA-By示例书中给出，以便在缓冲区中一个接一个地存储2个图像。如果我试图避免无限循环，我该如何编写程序，但我不确定我是否成功了？任何帮助编写正确的程序都将非常感谢！
#include "book.h"
#include "cpu_bitmap.h"
#include "cuda.h"
#include <cuda_gl_interop.h>
PFNGLBINDBUFFERARBPROC glBindBuffer = NULL;
PFNGLDELETEBUFFERSARBPROC glDeleteBuffers = NULL;
PFNGLGENBUFFERSARBPROC glGenBuffers = NULL;
PFNGLBUFFERDATAARBPROC glBufferData = NULL;
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, char *file, int line, bool abort=true)
{
if (code != cudaSuccess)
{
fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) system ("pause");
}
}
#define DIM 512
#define IMAGESIZE_MAX (DIM*DIM)
GLuint bufferObj;
cudaGraphicsResource *resource;
// based on ripple code, but uses uchar4 which is the type of data
// graphic inter op uses. see screenshot - basic2.png
__global__ void kernel( uchar4 *ptr1)
{
// map from threadIdx/BlockIdx to pixel position
int x = threadIdx.x + blockIdx.x * blockDim.x;
int y = threadIdx.y + blockIdx.y * blockDim.y;
int offset = x + y * blockDim.x * gridDim.x ;
// now calculate the value at that position
float fx = x/(float)DIM - 0.5f;
float fy = y/(float)DIM - 0.5f;
unsigned char green = 128 + 127 * tan( abs(fx*100) - abs(fy*100) );
// accessing uchar4 vs unsigned char*
ptr1[offset].x = 0;
ptr1[offset].y = green;
ptr1[offset].z = 0;
ptr1[offset].w = 255;
}
__global__ void kernel2( uchar4 *ptr2)
{
// map from threadIdx/BlockIdx to pixel position
int x = threadIdx.x + blockIdx.x * blockDim.x;
int y = threadIdx.y + blockIdx.y * blockDim.y;
int offset = x + y * blockDim.x * gridDim.x ;
// now calculate the value at that position
float fx = x/(float)DIM - 0.5f;
float fy = y/(float)DIM - 0.5f;
unsigned char green = 128 + 127 * tan( abs(fx*100) - abs(fy*100) );
unsigned char orange = 1000;
// accessing uchar4 vs unsigned char*
ptr2[offset].x = orange;
ptr2[offset].y = green;
ptr2[offset].z = 0;
ptr2[offset].w = 255;
}
__global__ void copy ( uchar4 *pBuffer, uchar4 *Ptr )
{
int x = threadIdx.x + blockIdx.x * blockDim.x;
int y = threadIdx.y + blockIdx.y * blockDim.y;
int idx = x + y * blockDim.x * gridDim.x ;
while ( idx != DIM*DIM)
{
pBuffer[idx] = Ptr[idx] ;
__syncthreads();
}
}
__global__ void copy2 ( uchar4 *pBuffer, uchar4 *Ptr2 )
{
int x = threadIdx.x + blockIdx.x * blockDim.x;
int y = threadIdx.y + blockIdx.y * blockDim.y;
int idx = x + y * blockDim.x * gridDim.x ;
int bdx = idx;
while ( (idx < DIM*DIM) && (bdx < DIM*DIM) )
{
uchar4 temp = Ptr2[bdx];
__syncthreads();
pBuffer[idx+4] = temp;
__syncthreads();
if ((idx==DIM*DIM) && (bdx==DIM*DIM))
{
break;
}
}
}
void key_func( unsigned char key, int x, int y ) {
switch (key) {
case 27:
// clean up OpenGL and CUDA
( cudaGraphicsUnregisterResource( resource ) );
glBindBuffer( GL_PIXEL_UNPACK_BUFFER_ARB, 0 );
glDeleteBuffers( 1, &bufferObj );
exit(0);
}
}
void draw_func( void ) {
// we pass zero as the last parameter, because out bufferObj is now
// the source, and the field switches from being a pointer to a
// bitmap to now mean an offset into a bitmap object
glDrawPixels( DIM, DIM, GL_RGBA, GL_UNSIGNED_BYTE, 0 );
glutSwapBuffers();
}
int main( int argc, char **argv ) {
cudaDeviceProp prop;
int dev;
(memset( &prop, 0, sizeof( cudaDeviceProp ) ));
prop.major = 1;
prop.minor = 0;
HANDLE_ERROR( cudaChooseDevice( &dev, &prop ) );
// tell CUDA which dev we will be using for graphic interop
// from the programming guide: Interoperability with OpenGL
// requires that the CUDA device be specified by
// cudaGLSetGLDevice() before any other runtime calls.
HANDLE_ERROR( cudaGLSetGLDevice( dev ) );
// these GLUT calls need to be made before the other OpenGL
// calls, else we get a seg fault
glutInit( &argc, argv );
glutInitDisplayMode( GLUT_DOUBLE | GLUT_RGBA );
glutInitWindowSize( DIM, DIM );
glutCreateWindow( "bitmap" );
glBindBuffer = (PFNGLBINDBUFFERARBPROC)GET_PROC_ADDRESS("glBindBuffer");
glDeleteBuffers = (PFNGLDELETEBUFFERSARBPROC)GET_PROC_ADDRESS("glDeleteBuffers");
glGenBuffers = (PFNGLGENBUFFERSARBPROC)GET_PROC_ADDRESS("glGenBuffers");
glBufferData = (PFNGLBUFFERDATAARBPROC)GET_PROC_ADDRESS("glBufferData");
// the first three are standard OpenGL, the 4th is the CUDA reg
// of the bitmap these calls exist starting in OpenGL 1.5
glGenBuffers( 1, &bufferObj );
glBindBuffer( GL_PIXEL_UNPACK_BUFFER_ARB, bufferObj );
glBufferData( GL_PIXEL_UNPACK_BUFFER_ARB, DIM * DIM * 4 ,
NULL, GL_DYNAMIC_DRAW_ARB );
// REGISTER THE GL BufferObj and CUDA Resource
HANDLE_ERROR(( cudaGraphicsGLRegisterBuffer( &resource,
bufferObj,
cudaGraphicsMapFlagsNone ) ));
// do work with the memory dst being on the GPU, gotten via mapping
HANDLE_ERROR( cudaGraphicsMapResources( 1, &resource, NULL ) );
uchar4* devPtr;
size_t size = DIM*DIM;
size_t sizet = 2*DIM*DIM;
gpuErrchk(cudaMalloc ( (uchar4 **)&devPtr, size));
uchar4 *devPtr2;
gpuErrchk(cudaMalloc ( (uchar4 **)&devPtr2, size));
uchar4 *pBuffer;
gpuErrchk(cudaMalloc ( (uchar4 **)&pBuffer, size));
uchar4 *pBufferCurrent;
gpuErrchk(cudaMalloc ( (uchar4 **)&pBuffer, size));
uchar4 *pBufferImage;
gpuErrchk(cudaMalloc ( (uchar4 **)&pBufferImage, sizet));
// REGISTER THE C BUFFER and CUDA Resource
HANDLE_ERROR( cudaGraphicsResourceGetMappedPointer( (void**)&pBufferImage,
&size,
resource) );
dim3 grids(DIM/16,DIM/16);
dim3 threads(16,16);
kernel<<<grids,threads>>>( devPtr );
gpuErrchk( cudaPeekAtLastError() );
gpuErrchk( cudaDeviceSynchronize() );
kernel2<<<grids,threads>>>(devPtr2);
gpuErrchk( cudaPeekAtLastError() );
gpuErrchk( cudaDeviceSynchronize() );
int a = 1;
do
{
if (a==1)
{
copy<<< 512, 512>>>(pBufferImage, devPtr);
gpuErrchk( cudaPeekAtLastError() );
gpuErrchk( cudaDeviceSynchronize() );
}
if(a==2)
{
copy2<<< 512, 512>>>(pBufferImage, devPtr2);
gpuErrchk( cudaPeekAtLastError() );
gpuErrchk( cudaDeviceSynchronize() );
}
a++;
} while (a<=2);
HANDLE_ERROR ( cudaGraphicsUnmapResources( 1, &resource, NULL ) );
// set up GLUT and kick off main loop
glutKeyboardFunc( key_func );
glutDisplayFunc( draw_func );
glutMainLoop();
}
答案 0 :(得分：0)
以下是我编写的一些代码，它是通过包含here的示例代码修改CUDA，我认为这些代码实际上就是您的开始。我使用两个内核，就像你一样，生成绿色或橙色图像。它最初将以显示的绿色图像开始，但您可以使用空格键在绿色和橙色图像之间切换。 ESC键将退出应用程序。
#include "book.h"
#include "cpu_bitmap.h"
//#include "cuda.h"
#include <cuda_gl_interop.h>
int which_image;
PFNGLBINDBUFFERARBPROC glBindBuffer = NULL;
PFNGLDELETEBUFFERSARBPROC glDeleteBuffers = NULL;
PFNGLGENBUFFERSARBPROC glGenBuffers = NULL;
PFNGLBUFFERDATAARBPROC glBufferData = NULL;
#define DIM 512
GLuint bufferObj;
cudaGraphicsResource *resource;
dim3 mgrids(DIM/16,DIM/16);
dim3 mthreads(16,16);
// based on ripple code, but uses uchar4 which is the type of data
// graphic inter op uses. see screenshot - basic2.png
__global__ void kernel_gr( uchar4 *ptr ) {
// map from threadIdx/BlockIdx to pixel position
int x = threadIdx.x + blockIdx.x * blockDim.x;
int y = threadIdx.y + blockIdx.y * blockDim.y;
int offset = x + y * blockDim.x * gridDim.x;
// now calculate the value at that position
float fx = x/(float)DIM - 0.5f;
float fy = y/(float)DIM - 0.5f;
unsigned char green = 128 + 127 *
sin( abs(fx*100) - abs(fy*100) );
// accessing uchar4 vs unsigned char*
ptr[offset].x = 0;
ptr[offset].y = green;
ptr[offset].z = 0;
ptr[offset].w = 255;
}
__global__ void kernel_or( uchar4 *ptr ) {
// map from threadIdx/BlockIdx to pixel position
int x = threadIdx.x + blockIdx.x * blockDim.x;
int y = threadIdx.y + blockIdx.y * blockDim.y;
int offset = x + y * blockDim.x * gridDim.x;
// now calculate the value at that position
float fx = x/(float)DIM - 0.5f;
float fy = y/(float)DIM - 0.5f;
unsigned char orange = 128 + 127 *
sin( abs(fx*100) - abs(fy*100) );
// accessing uchar4 vs unsigned char*
ptr[offset].x = orange;
ptr[offset].y = orange/2;
ptr[offset].z = 0;
ptr[offset].w = 255;
}
static void draw_func( void ) {
// we pass zero as the last parameter, because out bufferObj is now
// the source, and the field switches from being a pointer to a
// bitmap to now mean an offset into a bitmap object
glDrawPixels( DIM, DIM, GL_RGBA, GL_UNSIGNED_BYTE, 0 );
glutSwapBuffers();
}
static void key_func( unsigned char key, int x, int y ) {
switch (key) {
case 32:
// do work with the memory dst being on the GPU, gotten via mapping
HANDLE_ERROR( cudaGraphicsMapResources( 1, &resource, NULL ) );
uchar4* devPtr;
size_t size;
HANDLE_ERROR(
cudaGraphicsResourceGetMappedPointer( (void**)&devPtr,
&size,
resource) );
if (which_image == 1){
kernel_or<<<mgrids,mthreads>>>( devPtr );
HANDLE_ERROR(cudaPeekAtLastError());
HANDLE_ERROR(cudaDeviceSynchronize());
printf("orange\n");
which_image = 2;
}
else {
kernel_gr<<<mgrids,mthreads>>>( devPtr );
HANDLE_ERROR(cudaPeekAtLastError());
HANDLE_ERROR(cudaDeviceSynchronize());
printf("green\n");
which_image = 1;
}
HANDLE_ERROR( cudaGraphicsUnmapResources( 1, &resource, NULL ) );
draw_func();
break;
case 27:
// clean up OpenGL and CUDA
HANDLE_ERROR( cudaGraphicsUnregisterResource( resource ) );
glBindBuffer( GL_PIXEL_UNPACK_BUFFER_ARB, 0 );
glDeleteBuffers( 1, &bufferObj );
exit(0);
}
}
int main( int argc, char **argv ) {
cudaDeviceProp prop;
int dev;
memset( &prop, 0, sizeof( cudaDeviceProp ) );
prop.major = 1;
prop.minor = 0;
HANDLE_ERROR( cudaChooseDevice( &dev, &prop ) );
// tell CUDA which dev we will be using for graphic interop
// from the programming guide: Interoperability with OpenGL
// requires that the CUDA device be specified by
// cudaGLSetGLDevice() before any other runtime calls.
HANDLE_ERROR( cudaGLSetGLDevice( dev ) );
// these GLUT calls need to be made before the other OpenGL
// calls, else we get a seg fault
glutInit( &argc, argv );
glutInitDisplayMode( GLUT_DOUBLE | GLUT_RGBA );
glutInitWindowSize( DIM, DIM );
glutCreateWindow( "bitmap" );
glBindBuffer = (PFNGLBINDBUFFERARBPROC)GET_PROC_ADDRESS("glBindBuffer");
glDeleteBuffers = (PFNGLDELETEBUFFERSARBPROC)GET_PROC_ADDRESS("glDeleteBuffers");
glGenBuffers = (PFNGLGENBUFFERSARBPROC)GET_PROC_ADDRESS("glGenBuffers");
glBufferData = (PFNGLBUFFERDATAARBPROC)GET_PROC_ADDRESS("glBufferData");
// the first three are standard OpenGL, the 4th is the CUDA reg
// of the bitmap these calls exist starting in OpenGL 1.5
glGenBuffers( 1, &bufferObj );
glBindBuffer( GL_PIXEL_UNPACK_BUFFER_ARB, bufferObj );
glBufferData( GL_PIXEL_UNPACK_BUFFER_ARB, DIM * DIM * 4,
NULL, GL_DYNAMIC_DRAW_ARB );
HANDLE_ERROR(
cudaGraphicsGLRegisterBuffer( &resource,
bufferObj,
cudaGraphicsMapFlagsNone ) );
// do work with the memory dst being on the GPU, gotten via mapping
HANDLE_ERROR( cudaGraphicsMapResources( 1, &resource, NULL ) );
uchar4* devPtr;
size_t size;
HANDLE_ERROR(
cudaGraphicsResourceGetMappedPointer( (void**)&devPtr,
&size,
resource) );
dim3 grids(DIM/16,DIM/16);
dim3 threads(16,16);
kernel_gr<<<grids,threads>>>( devPtr );
HANDLE_ERROR( cudaGraphicsUnmapResources( 1, &resource, NULL ) );
which_image = 1;
// set up GLUT and kick off main loop
glutKeyboardFunc( key_func );
glutDisplayFunc( draw_func );
glutMainLoop();
}
不确定它是否有用，我仍然不明白你想完成什么。我真的不知道这意味着什么：
我只想将这些图像存储在缓冲区中，然后在OpenGL中渲染包含这两个图像的缓冲区。
您希望一次能看到一个图像，并切换图像？或者您希望能够同时看到两个图像？如果是后者，请解释。你想要一个在窗口顶部，一个在窗口底部吗？他们俩混在一起？
编辑：在我看来，您可能想要对多个图像进行某种3D可视化，因为您对所需内容的问答并没有提高效率（至少我仍然无法处理你想要看到的内容 VISUALLY ，忽略了幕后发生的事情。）你没有用OpenGL标记这个问题，所以没有OpenGL专家在看它。此外，您已经做了如下声明：“我将使用OpenGL函数来旋转和转换缓冲区。”如果您要做的是创建用户可以与之交互的一组图像的3D可视化，这不是您想要的示例代码。这是一个基本的2D图像显示代码。尝试扩展缓冲区以容纳多个图像是您在OpenGL中创建某种3D可视化的最小困难。并且您将无法使用此示例代码进行某种3D多图像显示。
我怀疑你尝试做的CUDA-OpenGL互操作部分并不困难。我已经在示例程序中展示了如何在用户控制下显示由2个不同内核生成的2个不同图像。因此，我认为如何从CUDA获取图像并显示它或将其放入可显示的缓冲区的问题就是很好的说明。
我的建议是：将CUDA-OpenGL互操作部分放在一边。使用任意图像编写一个OpenGL程序（根据需要生成它们，不需要使用CUDA。）如果需要帮助，请在SO上提出问题，并使用OpenGL 标记它们以便知道如何做的人可以帮助你。然后，当您有可视化显示的原型时，可以注入CUDA部分。而且我怀疑那部分会非常简单。