c - Handling Image pointer to pointer in CUDA -


I am trying to establish a CUDA code (part of the Fraclet Image Compression by Equal Fisher) After taking care of the indicators to assign a double pointer indicator indicator to a 2D image, I still can not say "Warning: Global Memory Location, Indicator can not tell", with warning Partitioning partition error is also happening. Here I am here I am here As I also post (my apologies to duplicate the code posted) 

  #include & lt; cuda.h & gt; # Include & lt; Stdio.h & gt; # Include & lt; Stdlib.h & gt; #define hsize 256 #define vsize 256 #define IMAGE_TYPE Unsigned four __global__ zero card (IMAGE_TYPE matrix [] [hsize], IMAGE_TYPE matrix out [] [hsig]) {int tid = threadIdx.x; Int bead = block idx.x; matrixout [quote] [tid] = matrixin [quote] [tid]; } int fatal (char * s) {fprintf (stderr, "% s \ n", s); Return 1; } #define matrix_allocate (matrix, hsiese, vs, type) {\ TYPE * imptr; \ int _i; \ matrix = (TYPE **) malloc ((vs) * size (TYPE *)); \ imptr = (TYPE * (no memory in matrix allocation); \ For (_i = 0; _i and lieutent; vs; ++ _ I, IMPTR + = HSIII) \ matrix [_i] = IMPTR; \} \ At MIT () {typed IMAGE_TYPE IMarray [Vs] [HSIC]; IMAGE_TYPE ** hin_image, ** hout_image; IMRRA * din_image, * dout_image; // For Allocated Host Matrix Matrix_Onlock (hin_image, hsize, vsize, IMAGE_TYPE) (Int j = 0; j & lt; hsiaj; j ++) hin_image [i] [j] = 'a'; Matrix_allocate (hout_image, for example i = 0; i & lt; vi; i ++); hsize, vs, IMAGE_TYPE) // Allocated device memory cudaMalloc (void **) and din_image, (vs * hsize) * sizeof (IMAGE_TYPE); cudaMalloc ((void **) & amp; dout_image, (vsize * hsize) * sizeof (IMAGE_TYPE); cudaMemcpy (din_image, hin_image, (vs * hsize) * size (IMAGE_TYPE), cudaMemcpyHostToDevice; dim 3 threads (HSIEEEE, 1,1); Dim 3 block (vs, 1,1); Giri  gt; din_image, dout_image; cudaMemcpy (hout_image, dout_image, (vsize * hsize) * sizeof (IMAGE_TYPE), cudaMemcpyDeviceToHost); For (int i = 0; i   
 I want to know what is wrong with the standard 2D access to the image inside the kernel function. Any help would be highly welcomed.   
 
 
 I'm not trying to allocate my complex metrics to plan. The purpose of my suggestion was to make things easier for you to simplify 1-line allocation.  
 Besides, I do not think you have understood the given example. This was a 3D example, and 2 subscripts in typeffs were typed in a 2D version with a single subscript.  
 In fact it does not have any CUDA, it rotates around the understanding of arrays and indicators.  
 These were major changes made to work on your code:  
  #include & lt; Stdio.h & gt; # Include & lt; Stdlib.h & gt; #define hsize 256 #define vsize 256 #define IMAGE_TYPE Unsigned four __global__ zero card (IMAGE_TYPE matrix [] [hsize], IMAGE_TYPE matrix out [] [hsig]) {int tid = threadIdx.x; Int bead = block idx.x; matrixout [quote] [tid] = matrixin [quote] [tid]; } int fatal (char * s) {fprintf (stderr, "% s \ n", s); Return 1; } Int main () {typedef IMAGE_TYPE IMarray [hsize]; IMRRA * hin_image, * hout_image; IMRRA * din_image, * dout_image; // Allocated memory hin_image = (IMRRE *) malloc (HSCE * vs * size (IMAGE_TYPE)); hout_image = (IMRRE *) Molok (HSCE * * vs * size (IMAGE_TYPE)); For (int i = 0; i & lt; vs; i ++) for (int j = 0; j and lt; hsice; j ++) hin_image [i] [j] = 'a'; // Allocated device memory cudaMalloc ((zero **) and din_image, (vs * hsize) * sizeof (IMAGE_TYPE)); cudaMalloc ((zero **) & dout_image, (vsize * hsize) * sizeof (IMAGE_TYPE)); cudaMemset (dout_image, 0, (vs * hsize) * size (IMAGE_TYPE)); CudaMemcpy (din_image, hin_image, (vs * hsize) * size (IMAGE_TYPE), cudaMemcpyHostToDevice); Dim 3 threads (HSIEEEE, 1,1); Dim 3 block (vs, 1,1); Fall & lt; & Lt; & Lt; Blocks, threads & gt; & Gt; & Gt; (din_image, dout_image); CudaMemcpy (hout_image, dout_image, (vsize * hsize) * sizeof (IMAGE_TYPE), cudaMemcpyDeviceToHost); For (int i = 0; i

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