Question

我试图将数组表示为cuda中的tex2D ....经过数小时的调试后，我注意到100万个元素中的19个被错误地复制到纹理中，意味着作为二进制数组，我得到了0不过是1。

void evolve_gpu( byte* h_in, byte* h_out)
{  

//int SIZE = N * N * N * N * sizeof( float );
cudaEvent_t start, stop;
size_t d_in_pitch;
size_t d_out_pitch;
int len = 1002;

checkCudaErrors( cudaEventCreate(&start) );
checkCudaErrors( cudaEventCreate(&stop) );

// Allocate the device input image array
unsigned char *d_in = NULL;
unsigned char *d_out = NULL;
checkCudaErrors(cudaMallocPitch(&d_in, &d_in_pitch, sizeof(unsigned char)*len, len));
checkCudaErrors(cudaMallocPitch(&d_out, &d_out_pitch, sizeof(unsigned char)*len, len));

// Copy the host input image  to the device memory
checkCudaErrors(cudaMemcpy2D(d_in, d_in_pitch, h_in, sizeof(unsigned char)*len
    , sizeof(unsigned char)*len, len, cudaMemcpyHostToDevice));




/**************************** TEXTURE CONFIGURATION ******************************/
cudaResourceDesc resDesc;
memset(&resDesc, 0, sizeof(resDesc));
resDesc.resType = cudaResourceTypePitch2D;
resDesc.res.pitch2D.devPtr = d_in;
resDesc.res.pitch2D.pitchInBytes =  d_in_pitch;
resDesc.res.pitch2D.width = len;
resDesc.res.pitch2D.height = len;
resDesc.res.pitch2D.desc = cudaCreateChannelDesc<unsigned char>();

cudaTextureDesc texDesc;
memset(&texDesc, 0, sizeof(texDesc));
texDesc.readMode =  cudaReadModeElementType;
texDesc.normalizedCoords=false;
texDesc.addressMode[0]=cudaAddressModeBorder;
texDesc.addressMode[1]=cudaAddressModeBorder;

cudaTextureObject_t tex;
cudaCreateTextureObject(&tex, &resDesc, &texDesc, NULL);
/*********************************************************************************/

checkCudaErrors( cudaEventRecord(start, NULL) );

// Launch the CUDA Kernel
dim3 block = dim3(THREADS_X, THREADS_Y);
dim3 grid = dim3((len+block.x-1)/block.x,(len+block.y-1)/block.y);//25*50
evolve_kernel<<<grid, block>>>( tex, d_out );

//******** kernel<<< number of blocks, number of threads, dynamic memory per block, associated stream >>>  *******//

// Copy the device result to the host
checkCudaErrors(cudaMemcpy2D(h_out, d_out_pitch,
                             d_out, d_out_pitch,
                             sizeof(unsigned char)*len,  len,
                             cudaMemcpyDeviceToHost));      
for(int i=0;i<1002*1002;i++){

    if(h_in[i] != h_out[i])
        printf("i = %d\n",i);


}
checkCudaErrors( cudaGetLastError() );

checkCudaErrors( cudaEventRecord(stop, NULL) );
checkCudaErrors( cudaEventSynchronize(stop) );

checkCudaErrors( cudaFree(d_in) );
checkCudaErrors( cudaFree(d_out) );

float msec = 0.f;
checkCudaErrors( cudaEventElapsedTime(&msec, start, stop) );

printf("Basic version took: %f ms\n", msec);

}

Answer 1

我可以在您的代码中看到的一个问题是您的设备 - ＆gt;主机副本：

checkCudaErrors(cudaMemcpy2D(h_out, d_out_pitch,
                         d_out, d_out_pitch,
                         sizeof(unsigned char)*len,  len,
                         cudaMemcpyDeviceToHost));

参考documentation，此cudaMemcpy2D来电的第二个参数是目的地分配的音高（即h_out的音高，在此案件）。但是h_out不太可能指的是分配，即使它不知何故，d_out_pitch也不太可能给出音调。

虽然您没有显示完整的代码，但假设h_out和h_in是类似的分配，那么第二个参数应该更改为{{1}的（无节距）宽度} array：

h_out

我还很好奇当你没有将checkCudaErrors(cudaMemcpy2D(h_out, len*sizeof(unsigned char), d_out, d_out_pitch, sizeof(unsigned char)*len, len, cudaMemcpyDeviceToHost));的音调传递给d_out时，你的内核如何在d_out（一个音调分配）上正常运行：

evolve_kernel<<<grid, block>>>( tex, d_out );

我原本希望看到这样的电话：

evolve_kernel<<<grid, block>>>( tex, d_out, d_out_pitch);

但你没有展示你的内核代码。

这是一个完整的示例，我围绕您展示的代码创建，修复了上述问题，并进行了一些其他更改以构建示例：

$ cat t648.cu
#include <stdio.h>
#include <helper_cuda.h>
#define THREADS_X 16
#define THREADS_Y 16

const int len = 1002;
typedef unsigned char byte;

__global__ void evolve_kernel(cudaTextureObject_t tex, unsigned char *d_out, size_t pitch ){
  int idx = threadIdx.x+blockDim.x*blockIdx.x;
  int idy = threadIdx.y+blockDim.y*blockIdx.y;
  if ((idx < len) && (idy < len))
    d_out[idy*pitch+idx] = tex2D<unsigned char>(tex, idx, idy);
}



void evolve_gpu( byte* h_in, byte* h_out)
{

//int SIZE = N * N * N * N * sizeof( float );
  cudaEvent_t start, stop;
  size_t d_in_pitch;
  size_t d_out_pitch;

  checkCudaErrors( cudaEventCreate(&start) );
  checkCudaErrors( cudaEventCreate(&stop) );

// Allocate the device input image array
  unsigned char *d_in = NULL;
  unsigned char *d_out = NULL;
  checkCudaErrors(cudaMallocPitch(&d_in, &d_in_pitch, sizeof(unsigned char)*len, len));
  checkCudaErrors(cudaMallocPitch(&d_out, &d_out_pitch, sizeof(unsigned char)*len, len));

// Copy the host input image  to the device memory
  checkCudaErrors(cudaMemcpy2D(d_in, d_in_pitch, h_in, sizeof(unsigned char)*len
    , sizeof(unsigned char)*len, len, cudaMemcpyHostToDevice));




/**************************** TEXTURE CONFIGURATION ******************************/
  cudaResourceDesc resDesc;
  memset(&resDesc, 0, sizeof(resDesc));
  resDesc.resType = cudaResourceTypePitch2D;
  resDesc.res.pitch2D.devPtr = d_in;
  resDesc.res.pitch2D.pitchInBytes =  d_in_pitch;
  resDesc.res.pitch2D.width = len;
  resDesc.res.pitch2D.height = len;
  resDesc.res.pitch2D.desc = cudaCreateChannelDesc<unsigned char>();

  cudaTextureDesc texDesc;
  memset(&texDesc, 0, sizeof(texDesc));
  texDesc.readMode =  cudaReadModeElementType;
  texDesc.normalizedCoords=false;
  texDesc.addressMode[0]=cudaAddressModeBorder;
  texDesc.addressMode[1]=cudaAddressModeBorder;

  cudaTextureObject_t tex;
  cudaCreateTextureObject(&tex, &resDesc, &texDesc, NULL);
/*********************************************************************************/

  checkCudaErrors( cudaEventRecord(start, NULL) );

// Launch the CUDA Kernel
  dim3 block = dim3(THREADS_X, THREADS_Y);
  dim3 grid = dim3((len+block.x-1)/block.x,(len+block.y-1)/block.y);//25*50
  evolve_kernel<<<grid, block>>>( tex, d_out, d_out_pitch );

//******** kernel<<< number of blocks, number of threads, dynamic memory per block, associated stream >>>  *******//

// Copy the device result to the host
  checkCudaErrors(cudaMemcpy2D(h_out, len*sizeof(unsigned char),
                             d_out, d_out_pitch,
                             sizeof(unsigned char)*len,  len,
                             cudaMemcpyDeviceToHost));
  for(int i=0;i<1002*1002;i++){

    if(h_in[i] != h_out[i])
        printf("i = %d\n",i);


  }
  checkCudaErrors( cudaGetLastError() );

  checkCudaErrors( cudaEventRecord(stop, NULL) );
  checkCudaErrors( cudaEventSynchronize(stop) );

  checkCudaErrors( cudaFree(d_in) );
  checkCudaErrors( cudaFree(d_out) );

  float msec = 0.f;
  checkCudaErrors( cudaEventElapsedTime(&msec, start, stop) );

  printf("Basic version took: %f ms\n", msec);

}

int main(){
  byte *h_data_in, *h_data_out;
  h_data_in  = (byte *)malloc(len*len*sizeof(byte));
  h_data_out = (byte *)malloc(len*len*sizeof(byte));
  for (int i = 0; i < len*len; i++){
    h_data_in[i]  = 3;
    h_data_out[i] = 0;}
  evolve_gpu(h_data_in, h_data_out);
  return 0;
}
$ nvcc -arch=sm_35 -I/usr/local/cuda/samples/common/inc t648.cu -o t648
$ ./t648
Basic version took: 3.868576 ms
$

它似乎正常工作并通过了您创建的测试。

将数组转换为纹理表示

1 个答案: