[QST] [CuTeDSL] `cast_tensor` corner case when no stride = 1

What is your question?

When none of a tensor's stride includes 1, cute.recast_tensor becomes no-op.

@cute.jit
def func():
    tensor = cute.make_rmem_tensor((1, 2), cute.Float64)
    print(tensor.shape, "->", cute.recast_tensor(tensor, dtype=cute.Float16).shape)  # --> works

    tensor = cute.make_rmem_tensor((1, 1), cute.Float64)
    print(tensor.shape, "->", cute.recast_tensor(tensor, dtype=cute.Float16).shape)  # --> no-op

    layout = cute.make_layout(
        shape=(((2, 2, 2), 1), 1, 2),
        stride=(((0, 16, 0), 0), 0, 0),
    )
    tensor = cute.make_rmem_tensor(layout, cute.Float64)
    print(tensor.shape, "->", cute.recast_tensor(tensor, dtype=cute.Float16).shape)  # --> no-op

    layout = cute.make_layout(
        shape=(((2, 2, 2), 1), 1, 2),
        stride=(((0, 1, 0), 0), 0, 0),
    )
    tensor = cute.make_rmem_tensor(layout, cute.Float64)
    print(tensor.shape, "->", cute.recast_tensor(tensor, dtype=cute.Float16).shape)  # --> works

func()

Start by running the provided cute.jit reproducer and inspect the cute.recast_tensor entry point, comparing the cases with and without a stride of 1. Done means the no-stride-1 cases have the intended recast shape and a regression check covers these examples.