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;
}

[download]
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;
}

[download]

See also:

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

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