Why does Torch computation operation call DtoH memcpy?

Here’s a code example.

import torch
import torch.cuda.nvtx as nvtx_cuda

A = torch.rand(size=(1024*4, 1024*4), device="cuda")
B = torch.rand(size=(1024*4, 1024*4), device="cuda")
C = torch.rand(size=(1024*4, 1024*4), device="cuda")
D = torch.rand(size=(1024*4, 1024*4), device="cuda")
E = torch.rand(size=(1024*4, 1024*4), device="cuda")
F = torch.rand(size=(1024*4, 1024*4), device="cuda")


idx_mask = torch.rand(1024, device="cuda")
active_idx = torch.rand(1024, device="cuda")
active_idx = torch.nonzero(idx_mask, as_tuple=True)

C = A.matmul(B)
F = D.matmul(E)



Regarding the code, I’m very curious about why active_idx = torch.nonzero(idx_mask, as_tuple=True) calls CudaMemcpyAsync.

Here’re my points.

  • idx_mask and active_idx are both on the GPU first.
  • So, all operations should be conducted on a GPU (independent of the CPU).
  • I suspect that the python torch.nonzero calls a CUDA kernel that does a non-zero operation in C++ within its underneath code because, in any way, the kernel should be called by the CPU thread. For example, GPU should send something like a pointer to the CPU, and the CPU launches the kernel.
  • So I make an example that calls a GEMM kernel, but that kernel does not call any memory operation (This also does the GEMM operation for tensors on a GPU).
  • In summary, what I want to do is eliminate the memory operation (GPU->CPU), making it independent of the CPU or at least make it to be a non-blocking operation which means eliminating the cudaSyncThread (the green block in the figure).

enter image description here
This is a non-zero operation.

enter image description here
This is a GEMM kernel.

Please note that the cudaMemasync is not called by the GEMM kernel in Figure 2.

Have you experimented to see what conditions produce/don’t produce the cudaMemcpyAsync call in S1 (e.g., removing the print statement / removing the matmul calls)?

torch.nonzero() causes a host-device synchronization as described in the docs as, I believe, the size of the output tensor has to be computed on the host first since its data-dependent.

1 Like

Thanks, I really appreciate it.