[QST] Question About Non-owning Tensor

[ZhangZhiPku]

In the tutorial:
https://github.com/NVIDIA/cutlass/blob/main/media/docs/cute/0y_predication.md

We have the following code snippet:

Tensor cA   = make_identity_tensor(make_shape(size<0>(sA), size<1>(sA)));  // (BLK_M,BLK_K) -> (blk_m,blk_k)
Tensor tAcA = local_partition(cA, tA, thread_idx);

Tensor cB   = make_identity_tensor(make_shape(size<0>(sB), size<1>(sB)));  // (BLK_N,BLK_K) -> (blk_n,blk_k)
Tensor tBcB = local_partition(cB, tB, thread_idx);

// Populate
CUTE_UNROLL
for (int m = 0; m < size<0>(tApA); ++m) {
  tApA(m,0) = get<0>(tAcA(m,0)) < m_max_coord;
}
CUTE_UNROLL
for (int n = 0; n < size<0>(tBpB); ++n) {
  tBpB(n,0) = get<0>(tBcB(n,0)) < n_max_coord;
}

In the above code, we created two predicate tensors, cA and cB.
I found that when calling the make_identity_tensor function, we are actually creating a tensor view(cA, cB). No memory is allocated to store the contents of cA or cB; instead, an iterator is created.( https://github.com/NVIDIA/cutlass/blob/cc3c29a81a140f7b97045718fb88eb0664c37bd7/include/cute/tensor_impl.hpp)

So, why are we able to modify cA, cB matrix in the tutorial code with:

tApA(m,0) = get<0>(tAcA(m,0)) < m_max_coord;
tBpB(n,0) = get<0>(tBcB(n,0)) < n_max_coord;

Does this call implicitly convert the cB matrix from a tensor view into a tensor with actual allocated memory?

[ccecka]

tAcA is the (tiled, sliced, partitioned) coordinate tensor. You're right that it is read-only.

tApA is the predicate tensor that actually stores bool

Tensor tApA = make_tensor<bool>(shape(tAcA));

which is read/write. So the above code is precomputing and storing predicates (to reuse in a mainloop, for example) via the read-only coordinate tensor.

I agree that I should revisit+update that documentation though...