I need to downsample an image.Some reading suggested me that if I use texture memory then this function comes on free and is faster(I am looking for a bilinear interpolation).Could some one tell me how exactly to write the kernel for this ?Here is what I have currently : (I use (1,1) thread blocks)
__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;
};
This version currently gives me all 0's in the output.
Edited (In this version I am trying to downsample 2 2000*512 size images into 2 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_开发者_开发知识库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 );
int id=blockIdx.x+256*blockIdx.y;
This statement crosses the limits of your final_red.
Try this:
__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);
}
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