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Fix unaligned access in SipHash-2-4.

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The compiler is allowed to assume that a "uint64_t *" is aligned
correctly, and will inline a version of memcpy that acts as such.

Use "uint8_t *", so the compiler does the right thing.
This commit is contained in:
Yawning Angel 2015-03-22 22:31:08 +00:00 committed by Nick Mathewson
parent 9063f29160
commit 732f522a42
2 changed files with 17 additions and 3 deletions
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4
changes/bug15436 Normal file
View File

@ -0,0 +1,4 @@
o Minor bugfixes (portability):
- Use the correct datatype in the SipHash-2-4 function to prevent compilers
from assuming any sort of alignment. Fixes bug 15436; bugfix on
0.2.5.3-alpha.

12
src/ext/csiphash.c
View File

@ -100,7 +100,15 @@ uint64_t siphash24(const void *src, unsigned long src_sz, const struct sipkey *k
uint64_t k0 = key->k0;
uint64_t k1 = key->k1;
uint64_t b = (uint64_t)src_sz << 56;
#ifdef UNALIGNED_OK
const uint64_t *in = (uint64_t*)src;
#else
/* On platforms where alignment matters, if 'in' is a pointer to a
* datatype that must be aligned, the compiler is allowed to
* generate code that assumes that it is aligned as such.
*/
const uint8_t *in = (uint8_t *)src;
#endif
uint64_t t;
uint8_t *pt, *m;
@ -113,12 +121,14 @@ uint64_t siphash24(const void *src, unsigned long src_sz, const struct sipkey *k
while (src_sz >= 8) {
#ifdef UNALIGNED_OK
uint64_t mi = _le64toh(*in);
in += 1;
#else
uint64_t mi;
memcpy(&mi, in, 8);
mi = _le64toh(mi);
in += 8;
#endif
in += 1; src_sz -= 8;
src_sz -= 8;
v3 ^= mi;
DOUBLE_ROUND(v0,v1,v2,v3);
v0 ^= mi;