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Re^4: The 10**21 Problem (Part 4)

by eyepopslikeamosquito (Canon)
on May 20, 2014 at 13:12 UTC ( #1086782=note: print w/ replies, xml ) Need Help??


in reply to Re^3: The 10**21 Problem (Part 4)
in thread The 10**21 Problem (Part 4)

Yes, I believe you are correct.

From stringobject.c:

static long string_hash(PyStringObject *a) { register Py_ssize_t len; register unsigned char *p; register long x; if (a->ob_shash != -1) return a->ob_shash; len = Py_SIZE(a); p = (unsigned char *) a->ob_sval; x = *p << 7; while (--len >= 0) x = (1000003*x) ^ *p++; x ^= Py_SIZE(a); if (x == -1) x = -2; a->ob_shash = x; return x; }
we can see that it is not whether the platform itself is 64-bit that matters, but whether the long type used by the C compiler that Python was built with is 64-bit. For Python built with a 32-bit long my solution should work, for a 64-bit long it will not.

On 64-bit architectures, Windows C compilers tend to use the LLP64 programming model (32-bit long), while most others tend to use the LP64 model (64-bit long). From this stack overflow question:

The true "war" was for sizeof(long), where Microsoft decided for sizeof(long) == 4 (LLP64) while nearly everyone else decided for sizeof(long) == 8 (LP64). Note that a programming model is a choice made on a per-compiler basis, and several can coexist on the same OS. However, the programming model chosen as the primary model for the OS API typically dominates.

Hmmm, I see from this later stringobject.c that _Py_HashSecret_* has been added, presumably to protect against DoS attacks that exploit hash collisions in Python dictionaries.

static long string_hash(PyStringObject *a) { register Py_ssize_t len; register unsigned char *p; register long x; #ifdef Py_DEBUG assert(_Py_HashSecret_Initialized); #endif if (a->ob_shash != -1) return a->ob_shash; len = Py_SIZE(a); /* We make the hash of the empty string be 0, rather than using (prefix ^ suffix), since this slightly obfuscates the hash secre +t */ if (len == 0) { a->ob_shash = 0; return 0; } p = (unsigned char *) a->ob_sval; x = _Py_HashSecret.prefix; x ^= *p << 7; while (--len >= 0) x = (1000003*x) ^ *p++; x ^= Py_SIZE(a); x ^= _Py_HashSecret.suffix; if (x == -1) x = -2; a->ob_shash = x; return x; }

See also:


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