+21
−29
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Compute a mask that will only have 0x80 in the bytes which
had a zero in them. The formula is:
~(((x & 0x7f7f7f7f) + 0x7f7f7f7f) | x | 0x7f7f7f7f)
In the inner word iteration, we have to compute the "x | 0x7f7f7f7f"
part, so we can reuse that in the above calculation.
Once we have this mask, we perform divide and conquer to find the
highest 0x80 location.
Signed-off-by:
David S. Miller <davem@davemloft.net>