It turns out that when locking a region, the region must be a naturally
aligned power of 2. The upshot of this is that if the desired region
crosses a 'large boundary' the region size must be increased
significantly to ensure that the locked region completely covers the
desired region. Previous calculations (including in kbase for the
proprietary driver) failed to take this into account.

Since it's known that the lock region must be naturally aligned we can
compute the required size by looking at the highest bit position which
changes between the start/end of the lock region (subtracting 1 from the
end because the end address is exclusive). The start address is then
aligned based on the size (this is technically unnecessary as the
hardware will ignore these bits, but the spec advises to do this "to
avoid confusion").
Reviewed-by: Boris Brezillon <boris.brezillon@collabora.com>
Reviewed-by: Alyssa Rosenzweig <alyssa.rosenzweig@collabora.com>
Signed-off-by: Steven Price <steven.price@arm.com>
Link: https://patchwork.freedesktop.org/patch/msgid/20210903094957.74560-1-steven.price@arm.com