Question

我需要对图像进行下采样。有些读数提示我，如果我使用纹理存储器，那么这个函数是免费的并且更快（我正在寻找双线性插值）。可以有人告诉我如何编写内核为此？这是我目前的情况：（我使用（1,1）线程块）

__global__ void texturekernel( int * final_red){
     int f = (blockIdx.x * blockDim.x) + threadIdx.x;
     int c = (blockIdx.y * blockDim.y) + threadIdx.y;
     int id=blockIdx.x+256*blockIdx.y;//256 is the width of downsampled image ..original was 512
     final_red[id]=tex2D( refTexture,c+0.5f,f+0.5f);//This is just for the red channel
     //where reftexture is defined as texture <float, 2, cudaReadModeElementType> refTexture;

  };

这个版本目前在输出中给我全部0。

已编辑（在此版本中，我尝试将2张2000 * 512尺寸的图片缩减为2张1000 * 256）：

  texture <float, 2, cudaReadModeElementType> refTexture; // global variable !
    cudaArray* myArray; 
    cudaChannelFormatDesc description = cudaCreateChannelDesc<float>();
    cudaError rs=cudaMallocArray (   &myArray,&description, 512,2000*2);//



//This line below is part of loop where input image is read row by row ..rowchecker keeps track of the row    

cudaMemcpyToArray(myArray,0,rowchecker++,array_temp_red,sizeof(int)*test_columns,cudaMemcpHostToDevice);

     refTexture.normalized=false;
     refTexture.addressMode[0]=cudaAddressModeClamp;
     refTexture.addressMode[1]=cudaAddressModeClamp;
     refTexture.filterMode=cudaFilterModePoint;

     cudaBindTextureToArray( refTexture,myArray);

              dim3 blockSize(1,1);
              int n_blocks_x=256;
              int n_blocks_y=1000*2;
              dim3 gridSize(n_blocks_x,n_blocks_y);
              cudaMalloc((void**)&finalarray,(2000)*(512)*2/4*sizeof(int));
              texturekernel<<<gridSize,blockSize>>>(finalarray );

Answer 1

int id = blockIdx.x + 256 * blockIdx.y;

此声明超越了final_red的限制。

试试这个：

__global__ void texturekernel( int * final_red){

     int f = blockIdx.x * blockDim.x + threadIdx.x;
     int c = blockIdx.y * blockDim.y + threadIdx.y;

     int id =c/2 * 256 + f/2;

     final_red[id] = tex2D( refTexture,c+0.5f,f+0.5f);

  }

使用纹理存储器对图像进行下采样

1 个答案: