[QST]why 3090 get different result with 4090 or 3060 when call get<0>(tensor)

I encountered a strange phenomenon.

for code below:

  auto l = make_identity_tensor(make_shape(_32{}, 64));

  using T_V = cutlass::AlignedArray<cutlass::half_t, 8>;
  using v_g2r_copy_op = UniversalCopy<T_V>;
  using v_g2r_traits = Copy_Traits<v_g2r_copy_op>;
  using v_g2r_copy_atom = Copy_Atom<v_g2r_traits, T_V>;

  using v_g2r_copy_tile = decltype(make_tiled_copy(
      v_g2r_copy_atom{}, Layout<Shape<_16, _8>, Stride<_8, _1>>{},
      Layout<Shape<_1, _8>>{}));

  v_g2r_copy_tile v_copy_tile;
  auto v_g2r_thr_copy = v_copy_tile.get_thread_slice(10);

  auto thr_v_c = v_g2r_thr_copy.partition_S(l);
  printf("get<0>(thr_v_c(0, 0, 0)) : %d\n", int(get<0>(thr_v_c(0, 0, 0))));
  printf("get<1>(thr_v_c(0, 0, 0)): %d\n", int(get<1>(thr_v_c(0, 0, 0))));
  bool can = get<0>(thr_v_c(0, 0, 0)) < 32 && get<1>(thr_v_c(0, 0, 0)) < 64;
  printf("can: %d\n", can);

when cutlass v3.0.0 , run code in cpu, GPU3060, 3090, 4090, it all get right result:

get<0>(thr_v_c(0, 0, 0)) : 1
get<1>(thr_v_c(0, 0, 0)): 16
can: 1

but when cutlass v3.5.0 , run code in cpu, GPU3060, 3090, 4090, only GPU3090 get wrong result:

get<0>(thr_v_c(0, 0, i)) : 0
get<1>(thr_v_c(0, 0, i)): 0
can: 0

Compilation directives: nvcc --std=c++17 -arch=sm_86 -w -I ./include/ test.cu -o a

Can you reproduce it?

Start by compiling and running the provided test.cu with the stated nvcc command on CUTLASS v3.0.0 and v3.5.0, comparing CPU, RTX 3060, RTX 3090, and RTX 4090 results. Trace make_identity_tensor, partition_S, and get<> to determine whether the RTX 3090 discrepancy is reproducible and identify the expected result for the reported case.