![]() For 1D and 2D grids and blocks, the unused dimensions should be set to 1 for clarity. ![]() Each such parameter is of type dim3, which is a C struct with three unsigned integer fields: x, y, and z. The second ECP specifies the dimensions of each block in number of threads. the mentioned example above, are doing more or less the same thing when dispatching the kernel.Īny help is appreciated, thanks in advance. The first ECP specifies the dimensions of the grid in number of blocks. I do not find any error, as other code, e.g. If I add cudaDeviceSynchronize(), an error 700 is thrown with illegal memory access when I try to print out the result in Python. No values are copied towards the output, but CUDA does not throw an error (Error Code 0 is printed) and I get all “Hello from CUDA If” printouts. To review, open the file in an editor that reveals hidden Unicode characters. The matrix has the same shape, dtype and is on the same CUDA device, but it is just zero. This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. The entire project and the setup.py install output are available here in this Github Gist.Īs described above, if I remove cudaDeviceSynchronize() from reduce_cuda(…) in reduce_cuda_kernel.cu I get in Python the zero matrix created with auto output = torch::zeros_like(matrix) Īs a result from the kernel call. 98 Linear address calculation for threads and submatrices. The kernel code is the following, it copies the data from a const 3d input matrix into a 3d output matrix: template Ĭonst torch::PackedTensorAccessor32 matrix,Ĭonst int y = blockIdx.y * blockDim.y threadIdx.y Ĭonst int x = blockIdx.x * blockDim.x threadIdx.x My issue is that in this simple example I either get the created zero matrix as a result in Python or, if after the kernel call cudaDeviceSynchronize() is added, an illegal memory access error (Code 700). You can prove this to yourself by switching to double arithmetic.Following the C /CUDA extension tutorial on the pytorch website and having a look at the linked source code I have created my own CUDA kernel which does not do something useful, but is done as a learning project. So small differences in the 5th or 6th decimal place may be attributable to order of operations differences when doing floating-point arithmetic. Your posted code will fail if GRID_SIZE and BLOCK_SIZE are different (for example, if GRID_SIZE were smaller than BLOCK_SIZE, cuda-memcheck will show illegal accesses, and if GRID_SIZE is larger than BLOCK_SIZE then this indexing error will result in blocks overwriting each other's values in the output array) because of this mixup between blockDim and gridDim.Īlso note that float operations typically only have around 5 decimal digits of precision. ![]() The gridDim variable gives the dimensions of the grid in terms of blocks - and this lines up exactly with how our results array is set up. Therefore the width of this array is given by gridDim.x (not blockDim). We want a simulated 2D index into an array that has width and height of GRID_SIZE consisting of one float quantity per point. The important change I made to your kernel code was in the output indexing: blocksum = aggregate If (result != (float)(GRID_SIZE*GRID_SIZE*BLOCK_SIZE*BLOCK_SIZE)) printf("mismatch, should be: %f, was: %f\n", (float)(GRID_SIZE*GRID_SIZE*BLOCK_SIZE*BLOCK_SIZE), result) For a 2D grid, the number of threads in X dimension is equal to block.x grid.x and in Y dimension equal to block.y grid.y. organize the computation into 2D blocks with TX threads in the x-direction. It means that the total number of threads in a dimension is equal to the product of grid size and block size in that dimension. CUDA uses the vector type dim3 for the dimension variables, gridDim and. ![]() ![]() Your calculations first so that each thread holds its resultĭx = fabs(s->conf - s->conf) ĭy = fabs(s->conf - s->conf) ĭz = fabs(s->conf - s->conf) įor (int i = 0 i < GRID_SIZE*GRID_SIZE i ) result = h_result Remember that grid size means the number of block in each dimension. Int j = blockDim.y*blockIdx.y threadIdx.y Int i = blockDim.x*blockIdx.x threadIdx.x Void add(frame* s, float L, float rc, float* blocksum) I'm not sure how to return the values of each block to the Host when using 2-dimensional grids. I'm trying to make a sum using the CUB reduction method. ![]()
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