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由 Joel Becker 提交于
In the calc_code_bit() function, we must find all powers of two beneath the code bit number, *after* it's shifted by those powers of two. This requires a loop to see where it ends up. We can optimize it by starting at its most significant bit. This shaves 32% off the time, for a total of 67.6% shaved off of the original, naive implementation. Signed-off-by: NJoel Becker <joel.becker@oracle.com> Signed-off-by: NMark Fasheh <mfasheh@suse.com>
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