[QST] The definition of zero stride in CUTE layout algebra

What is your question?

auto sA_layout = make_layout(make_shape(bM, bK));                 // (m,k) -> smem_idx; m-major  
ThrCopy thr_copy_a = copy_a.get_slice(threadIdx.x); 
Tensor sA = make_tensor(make_smem_ptr(smemA), sA_layout);            // (BLK_M,BLK_K) 
Tensor tAsA = thr_copy_a.partition_D(sA);    

From the above snippet from sgemm_2.cu, I got the following from partitioning sA via ThrCopy:

sA:
smem_ptr[32b](0x6ffff4000000) o (_128,_8):(_1,_128)

tAsA:
smem_ptr[32b](0x6ffff4000ff0) o ((_4,_1),_1,_1):((_1,_0),_0,_0)

How to interpret those _0 in tAsA? In the tutorial, there was a mentioning that _0 in the dimensions of tAsA indicates a dimension with size 1 where the exact stride is not crucial, allowing it to be any value. But if we follow the coordinate-to-index mapping equation (index = shapes .* strides), _0 could also be interpreted as that the tensor is being broadcasted along that dimension?

Will the meaning of _0 change depending on the print scenario or it always means ignore the stride in this dimension?

Start with the cited sgemm_2.cu snippet and the tutorial’s discussion of _0; inspect the printed sA and tAsA layouts and the partition_D call. Done means the repository documentation or issue response clearly states how zero strides are interpreted and whether that interpretation depends on the printing context.