The memmove span covers from (next+1) to the end of the array, and the
index of next is (i+1), so the index of (next+1) is (i+2).  So the size
of remaining array elements is (type->cnt - (i + 2)).

Since the remaining elements of the memblock array are move forward by
one element and there is only one additional element caused by this bug.
So there won't be any write overflow here but read overflow.  It may
read one more element out of the array address if the array happens to
be full.  Commonly it doesn't matter at all but if the array happens to
be located at the end a memblock, it may cause a invalid read operation
for the physical address doesn't exist.

There are 2 *happens to be* here, so I think the probability is quite
low, I don't know if any guy is haunted by this bug before.

Mostly I think it's user-invisible.
Signed-off-by: Lin Feng <linfeng@cn.fujitsu.com>
Acked-by: Tejun Heo <tj@kernel.org>
Reviewed-by: Wanpeng Li <liwanp@linux.vnet.ibm.com>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>