Unsafe: efficiently comparing two byte arrays

Florian Weimer fweimer at redhat.com
Thu Feb 27 14:09:27 UTC 2014

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:


Florian Weimer / Red Hat Product Security Team

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