maple_tree: i is always less than or equal to mas_end

Patch series "refine mas_mab_cp()".

By analysis of the code, one condition check can be removed and one case
would hit a redundant assignment.


This patch (of 2):

mas_mab_cp() copy range [mas_start, mas_end] inclusively from a
maple_node to maple_big_node. This implies mas_start <= mas_end.

Based on the relationship of mas_start and mas_end, we can have the
following four cases:

                 | mas_start == mas_end |  mas_start < mas_end
  ---------------+----------------------+----------------------
  mas_start == 0 |         1            |          2
  ---------------+----------------------+----------------------
  mas_start != 0 |         3            |          4

We can see in all these four cases, i is always less than or equal to
mas_end after finish the loop:

  Case 1: After assign pivot 0, i is set to 1, which is bigger than
          mas_end 0. So it jumps to complete and skip the check.
  Case 2: After assign pivot 0, i is set to 1.
          ∵ (mas_start < mas_end) && (mas_start == 0)
             ==>  (1 <= mas_end)
          ∵ (i == 1) && (1 <= mas_end)
             ==>  (i <= mas_end)
          ∴ Before loop, we have (i <= mas_end). And we still hold this
             if it skips the loop. For example, (i == mas_end).

          Now let's see what happens in the loop:
          ∵ piv_end = min(mas_end, mt_pivots[mt])
             ==>  (piv_end <= mas_end)
	  ∵ loop condition is (i < piv_end)
	     ==>  (i <= piv_end) on finish the loop both normally or break
          ∵ (i <= piv_end) && (piv_end <= mas_end)
             ==>  (i <= mas_end)
          ∴ After loop, we still get (i <= mas_end) in this case
  Case 3: This case would skip both if clause and loop. So when it comes
          to the check, i is still mas_start which equals to mas_end.
  Case 4: This case would skip the if clause.
          ∵ (mas_start < mas_end) && (i == mas_start)
             ==>  (i < mas_end)
          ∴ Before loop, we have (i < mas_end).
          The loop process is similar with Case 2, so we get the same
	  result.

Now we can conclude in all cases, we get (i <= mas_end) when doing
check. Then it is not necessary to do the check.

Link: https://lkml.kernel.org/r/[email protected]
Link: https://lkml.kernel.org/r/[email protected]
Signed-off-by: Wei Yang <[email protected]>
Reviewed-by: Liam R. Howlett <[email protected]>
Signed-off-by: Andrew Morton <[email protected]>

0 files changed, 0 insertions, 0 deletions