【发布时间】：2019-09-29 05:29:01
【问题描述】：
我得到了以下程序，它几乎是 SDK 示例“简单分层纹理”。
// includes, system
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
// includes, kernels
#include <cuda_runtime.h>
// includes, project
#include <helper_cuda.h>
#include <helper_functions.h> // helper for shared that are common to CUDA SDK samples
#define EXIT_WAIVED 2
static char *sSDKname = "simpleLayeredTexture";
// includes, kernels
// declare texture reference for layered 2D float texture
// Note: The "dim" field in the texture reference template is now deprecated.
// Instead, please use a texture type macro such as cudaTextureType1D, etc.
typedef int TYPE;
texture<TYPE, cudaTextureType2DLayered> tex;
////////////////////////////////////////////////////////////////////////////////
//! Transform a layer of a layered 2D texture using texture lookups
//! @param g_odata output data in global memory
////////////////////////////////////////////////////////////////////////////////
__global__ void
transformKernel(TYPE *g_odata, int width, int height, int layer)
{
// calculate this thread's data point
unsigned int x = blockIdx.x*blockDim.x + threadIdx.x;
unsigned int y = blockIdx.y*blockDim.y + threadIdx.y;
// 0.5f offset and division are necessary to access the original data points
// in the texture (such that bilinear interpolation will not be activated).
// For details, see also CUDA Programming Guide, Appendix D
float u = (x+0.5f) / (float) width;
float v = (y+0.5f) / (float) height;
// read from texture, do expected transformation and write to global memory
TYPE sample = tex2DLayered(tex, u, v, layer);
g_odata[layer*width*height + y*width + x] = sample;
printf("Sample %d\n", sample);
}
////////////////////////////////////////////////////////////////////////////////
// Program main
////////////////////////////////////////////////////////////////////////////////
int
main(int argc, char **argv)
{
printf("[%s] - Starting...\n", sSDKname);
// use command-line specified CUDA device, otherwise use device with highest Gflops/s
int devID = findCudaDevice(argc, (const char **)argv);
bool bResult = true;
// get number of SMs on this GPU
cudaDeviceProp deviceProps;
checkCudaErrors(cudaGetDeviceProperties(&deviceProps, devID));
printf("CUDA device [%s] has %d Multi-Processors ", deviceProps.name, deviceProps.multiProcessorCount);
printf("SM %d.%d\n", deviceProps.major, deviceProps.minor);
if (deviceProps.major < 2)
{
printf("%s requires SM >= 2.0 to support Texture Arrays. Test will be waived... \n", sSDKname);
cudaDeviceReset();
exit(EXIT_SUCCESS);
}
// generate input data for layered texture
unsigned int width=16, height=16, num_layers = 5;
unsigned int size = width * height * num_layers * sizeof(TYPE);
TYPE *h_data = (TYPE *) malloc(size);
for (unsigned int layer = 0; layer < num_layers; layer++)
for (int i = 0; i < (int)(width * height); i++)
{
h_data[layer*width*height + i] = 15;//(float)i;
}
// this is the expected transformation of the input data (the expected output)
TYPE *h_data_ref = (TYPE *) malloc(size);
for (unsigned int layer = 0; layer < num_layers; layer++)
for (int i = 0; i < (int)(width * height); i++)
{
h_data_ref[layer*width*height + i] = h_data[layer*width*height + i];
}
// allocate device memory for result
TYPE *d_data = NULL;
checkCudaErrors(cudaMalloc((void **) &d_data, size));
// allocate array and copy image data
cudaChannelFormatDesc channelDesc = cudaCreateChannelDesc<TYPE>();
cudaArray *cu_3darray;
checkCudaErrors(cudaMalloc3DArray(&cu_3darray, &channelDesc, make_cudaExtent(width, height, num_layers), cudaArrayLayered));
cudaMemcpy3DParms myparms = {0};
myparms.srcPos = make_cudaPos(0,0,0);
myparms.dstPos = make_cudaPos(0,0,0);
myparms.srcPtr = make_cudaPitchedPtr(h_data, width * sizeof(TYPE), width, height);
myparms.dstArray = cu_3darray;
myparms.extent = make_cudaExtent(width, height, num_layers);
myparms.kind = cudaMemcpyHostToDevice;
checkCudaErrors(cudaMemcpy3D(&myparms));
// set texture parameters
tex.addressMode[0] = cudaAddressModeWrap;
tex.addressMode[1] = cudaAddressModeWrap;
// tex.filterMode = cudaFilterModeLinear;
tex.filterMode = cudaFilterModePoint;
tex.normalized = true; // access with normalized texture coordinates
// Bind the array to the texture
checkCudaErrors(cudaBindTextureToArray(tex, cu_3darray, channelDesc));
dim3 dimBlock(8, 8, 1);
dim3 dimGrid(width / dimBlock.x, height / dimBlock.y, 1);
printf("Covering 2D data array of %d x %d: Grid size is %d x %d, each block has 8 x 8 threads\n",
width, height, dimGrid.x, dimGrid.y);
transformKernel<<< dimGrid, dimBlock >>>(d_data, width, height, 0); // warmup (for better timing)
// check if kernel execution generated an error
getLastCudaError("warmup Kernel execution failed");
checkCudaErrors(cudaDeviceSynchronize());
StopWatchInterface *timer = NULL;
sdkCreateTimer(&timer);
sdkStartTimer(&timer);
// execute the kernel
for (unsigned int layer = 0; layer < num_layers; layer++)
transformKernel<<< dimGrid, dimBlock, 0 >>>(d_data, width, height, layer);
// check if kernel execution generated an error
getLastCudaError("Kernel execution failed");
checkCudaErrors(cudaDeviceSynchronize());
sdkStopTimer(&timer);
printf("Processing time: %.3f msec\n", sdkGetTimerValue(&timer));
printf("%.2f Mtexlookups/sec\n", (width *height *num_layers / (sdkGetTimerValue(&timer) / 1000.0f) / 1e6));
sdkDeleteTimer(&timer);
// allocate mem for the result on host side
TYPE *h_odata = (TYPE *) malloc(size);
// copy result from device to host
checkCudaErrors(cudaMemcpy(h_odata, d_data, size, cudaMemcpyDeviceToHost));
printf("Comparing kernel output to expected data\n");
#define MIN_EPSILON_ERROR 5e-3f
bResult = compareData(h_odata, h_data_ref, width*height*num_layers, MIN_EPSILON_ERROR, 0.0f);
printf("Host sample: %d == %d\n", h_data_ref[0], h_odata[0]);
// cleanup memory
free(h_data);
free(h_data_ref);
free(h_odata);
checkCudaErrors(cudaFree(d_data));
checkCudaErrors(cudaFreeArray(cu_3darray));
cudaDeviceReset();
if (bResult)
printf("Success!");
else
printf("Failure!");
exit(bResult ? EXIT_SUCCESS : EXIT_FAILURE);
}
如果我使用 int（或 uint）作为 TYPE，输出是正确的。对于 float 它会产生错误的结果，即始终为 0（尽管 SDK compareData 函数说一切都很好！？）。我开始相信 CUDA 中存在错误。我在 Kepler K20 上使用 5.0 版。
感谢任何建议和测试结果。代码应该可以按原样运行。
提前致谢， 本
编辑：操作系统是 Linux (Ubuntu 12.04.2 LTS) x86_64 3.2.0-38-generic
【问题讨论】：
标签： cuda