Question

我编写了下面的代码，看看如何为1D数组使用纹理内存。但是tex1D函数没有从数组中获取相应线程id的值。请更正此代码并告诉我如何有效地使用1D数组的纹理内存有效。

__global__ void sum(float *b,cudaTextureObject_t texObj)

    {
    b[threadIdx.x]=tex1D<float>(texObj,threadIdx.x);
    //printf("\n%f\n",tex1Dfetch<float>(texObj,threadIdx.x));
    }
    int main()
    {
    float *a,*b;
    float *d_a,*d_b;
    int i;
    a=(float*)malloc(sizeof(float)*5);
    b=(float*)malloc(sizeof(float)*5);

    for(i=0;i<5;i++)
        a[i]=i;

    cudaChannelFormatDesc channelDesc =cudaCreateChannelDesc(32, 0, 0, 0,cudaChannelFormatKindFloat);

    cudaArray* cuArray;
    cudaMallocArray(&cuArray, &channelDesc, 5, 0);

    cudaMemcpyToArray(cuArray, 0, 0, a,sizeof(float)*5,cudaMemcpyHostToDevice);


    struct cudaResourceDesc resDesc;
        memset(&resDesc, 0, sizeof(resDesc));
        resDesc.resType = cudaResourceTypeArray;
        resDesc.res.array.array = cuArray;


      struct cudaTextureDesc texDesc;
        memset(&texDesc, 0, sizeof(texDesc));
        texDesc.addressMode[0]   = cudaAddressModeWrap;
        texDesc.addressMode[1]   = cudaAddressModeWrap;
        texDesc.filterMode       = cudaFilterModeLinear;
        texDesc.readMode         = cudaReadModeElementType;
        texDesc.normalizedCoords = 1;

        // Create texture object
        cudaTextureObject_t texObj = 0;
        cudaCreateTextureObject(&texObj, &resDesc, &texDesc, NULL);


    cudaMalloc(&d_b, 5* sizeof(float));

    sum<<<1,5>>>(d_b,texObj);



        // Free device memory
    cudaMemcpy(b,d_b,sizeof(float),cudaMemcpyDeviceToHost);

     for(i=0;i<5;i++)
        printf("%f\t",b[i]);
      cudaDestroyTextureObject(texObj); 
    cudaFreeArray(cuArray);
    cudaFree(d_b);

        return 0;

    }

Answer 1

至少有两个问题：

您只是在最后将一个浮动数量从设备复制回主机：

cudaMemcpy(b,d_b,sizeof(float),cudaMemcpyDeviceToHost);
                 ^^^^^^^^^^^^^

如果要打印5个值，则应复制5个值：

cudaMemcpy(b,d_b,5*sizeof(float),cudaMemcpyDeviceToHost);

您选择了normalized coordinates：

texDesc.normalizedCoords = 1;

这意味着您应该将0到1之间的浮点坐标作为索引，而不是0到4之间的整数坐标：

 b[threadIdx.x]=tex1D<float>(texObj,threadIdx.x);
                                    ^^^^^^^^^^^

使用类似的东西：

 b[threadIdx.x]=tex1D<float>(texObj, ((float)threadIdx.x/5.0f));

$ cat t3.cu
#include <stdio.h>

__global__ void sum(float *b,cudaTextureObject_t texObj)

    {
    b[threadIdx.x]=tex1D<float>(texObj,((float)(threadIdx.x+1)/5.0f));

    //printf("\n%f\n",tex1Dfetch<float>(texObj,threadIdx.x));
    }


int main()
    {
    float *a,*b;
    float *d_b;
    int i;
    a=(float*)malloc(sizeof(float)*5);
    b=(float*)malloc(sizeof(float)*5);

    for(i=0;i<5;i++)
        a[i]=i;

    cudaChannelFormatDesc channelDesc =cudaCreateChannelDesc(32, 0, 0, 0,cudaChannelFormatKindFloat);

    cudaArray* cuArray;
    cudaMallocArray(&cuArray, &channelDesc, 5, 0);

    cudaMemcpyToArray(cuArray, 0, 0, a,sizeof(float)*5,cudaMemcpyHostToDevice);


    struct cudaResourceDesc resDesc;
        memset(&resDesc, 0, sizeof(resDesc));
        resDesc.resType = cudaResourceTypeArray;
        resDesc.res.array.array = cuArray;


      struct cudaTextureDesc texDesc;
        memset(&texDesc, 0, sizeof(texDesc));
        texDesc.addressMode[0]   = cudaAddressModeWrap;
        texDesc.addressMode[1]   = cudaAddressModeWrap;
        texDesc.filterMode       = cudaFilterModeLinear;
        texDesc.readMode         = cudaReadModeElementType;
        texDesc.normalizedCoords = 1;

        // Create texture object
        cudaTextureObject_t texObj = 0;
        cudaCreateTextureObject(&texObj, &resDesc, &texDesc, NULL);


    cudaMalloc(&d_b, 5* sizeof(float));

    sum<<<1,4>>>(d_b,texObj);



        // Free device memory
    cudaMemcpy(b,d_b,5*sizeof(float),cudaMemcpyDeviceToHost);

     for(i=0;i<4;i++)
        printf("%f\t",b[i]);
      printf("\n");
      cudaDestroyTextureObject(texObj);
    cudaFreeArray(cuArray);
    cudaFree(d_b);

        return 0;

    }
$ nvcc -arch=sm_61 -o t3 t3.cu
$ cuda-memcheck ./t3
========= CUDA-MEMCHECK
0.500000        1.500000        2.500000        3.500000
========= ERROR SUMMARY: 0 errors
$

请注意，我确实做了一些其他更改。特别是，我调整了您的采样点以及样本数量，以选择在您拥有的5个数据点（0,1,2,3,4）中间线性插值的样本点，从而产生总输出4个数量（0.5,1.5,2.5,3.5）代表5个数据点之间的中点。

如果您想了解有关规范化坐标索引的更多信息，请参阅the programming guide以及边框模式等其他概念。此外，有各种CUDA sample codes证明正确使用纹理。

如何在CUDA中为1D数组使用纹理内存

1 个答案: