On 02/27/2014 11:37 AM, Paul Sandoz wrote:
/*
* We want to compare only the first index where left[index] !=
right[index].
* This corresponds to the least significant nonzero byte in lw ^
rw, since lw
* and rw are little-endian. Long.numberOfTrailingZeros(diff)
tells us the least
* significant nonzero bit, and zeroing out the first three bits
of L.nTZ gives us the
* shift to get that least significant nonzero byte.
*/
int n = Long.numberOfTrailingZeros(lw ^ rw) & ~0x7;
return (int) (((lw >>> n) & UNSIGNED_MASK) - ((rw >>> n) &
UNSIGNED_MASK));
This really depends on the microarchitecture, but for many current CPUs,
conversion to big endian with Long::reverseBytes(long) will be faster
(assuming that it's intrinsified).
This is not just a performance optimization, it can also be used to plug
an alleged timing oracle:
<https://sourceware.org/ml/libc-alpha/2014-01/msg00763.html>
--
Florian Weimer / Red Hat Product Security Team
