![]() It is really easy to ignore these tiny differences. Threads = blocks=blocks threads=threads kernel_matmul(Cd', Ad', Bd', m, n, C = Cd M, n, p = 2, 2, 2 # matrix sizes: C = A * Bĭim3 dimGrid((numCColumns dimBlock.x-1)/dimBlock.x,(numCRows dimBlock.y-1)/dimBlock.y,1) Therefore, the working code should be: # Matrix multiplication in GPU JuliaĬvalue = A*B The key issues are indexing base and order:Ĭuda is row major ordered while Julia is column major ordered. Hi, I was trying to print out debugging information from within the thread, but no information was printed out if I used dim3 type to specify threads per block. Thanks for the useful information! I checked the code more carefully, and finally found the bug. Run Julia on debug level 2 for device stack traces. However, it showed me the error: ERROR: a exception was thrown during kernel execution. Pro Get powerful tools for managing your contents. Presentation Creator Create stunning presentation online in just 3 steps. Presentation Survey Quiz Lead-form E-Book. Threads = blocks=blocks threads=threads kernel_matmul(Cd, Ad, Bd, m, n, C = Cd Recent Presentations Content Topics Updated Contents Featured Contents. M, n, p = 8, 8, 8 # matrix sizes: C = A * BĪ, B, C = rand(precision,m,p), rand(precision,p,n), rand(precision,m,n)Īd, Bd, Cd = CuArray(A), CuArray(B), CuArray(C) # Write the matrix to device memory each thread writes one element Ty = (blockIdx().y-1) * blockDim().y threadIdx().y ![]() Tx = (blockIdx().x-1) * blockDim().x threadIdx().x Then I tried the simple naïve matrix multiplication in CUDA and converted it into Julia: # Matrix multiplication in GPU Julia From the limited tutorials, I notice that it is possible to write cuda kernels in Julia directly. Printf("CUDA error(8): %s \n", cudaGetErrorString(err)) ĭoes anybody know if there is a method of debugging a cuda application starting from v3.I am learning to use GPU in Julia. print_error(cudaGetErrorString(cudaGetLastError())) Įrr = cudaMemcpy(gpu_memory_block, cpu_memory_block, memSize, cudaMemcpyDeviceToHost) <= Same error occurs at this point as well Printf("CUDA error(7): %s \n", cudaGetErrorString(err)) Printf("CUDA error(6): %s \n", cudaGetErrorString(err)) Įrr = cudaMemcpy(gpu_found_index, cpu_found_index, foundSize, cudaMemcpyDeviceToHost) <= The error occurs here when trying to move mem from d to h Printf("CUDA error(5): %s \n", cudaGetErrorString(err)) Printf("CUDA error(4): %s \n", cudaGetErrorString(err)) Ĭompute_hashes_on_memory_block_items>(gpu_found_index, gpu_memory_block) // kernel executes correctly Printf("CUDA error(3): %s \n", cudaGetErrorString(err)) Įrr = cudaMemcpy(gpu_found_index, cpu_found_index, foundSize, cudaMemcpyHostToDevice) Printf("CUDA error(2): %s \n", cudaGetErrorString(err)) Įrr = cudaMemcpy(gpu_memory_block, (uint4*)cpu_memory_block, memSize, cudaMemcpyHostToDevice) // is this cast meaningful or not? Printf("CUDA error(1): %s \n", cudaGetErrorString(err)) Įrr = cudaMalloc( (void**)
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