Issue #8188: Introduce a new scheme for computing hashes of numbers

(instances of int, float, complex, decimal.Decimal and
fractions.Fraction) that makes it easy to maintain the invariant that
hash(x) == hash(y) whenever x and y have equal value.
This commit is contained in:
Mark Dickinson 2010-05-23 13:33:13 +00:00
parent 03721133a6
commit dc787d2055
14 changed files with 566 additions and 137 deletions

View file

@ -403,12 +403,12 @@ complex_str(PyComplexObject *v)
static long
complex_hash(PyComplexObject *v)
{
long hashreal, hashimag, combined;
hashreal = _Py_HashDouble(v->cval.real);
if (hashreal == -1)
unsigned long hashreal, hashimag, combined;
hashreal = (unsigned long)_Py_HashDouble(v->cval.real);
if (hashreal == (unsigned long)-1)
return -1;
hashimag = _Py_HashDouble(v->cval.imag);
if (hashimag == -1)
hashimag = (unsigned long)_Py_HashDouble(v->cval.imag);
if (hashimag == (unsigned long)-1)
return -1;
/* Note: if the imaginary part is 0, hashimag is 0 now,
* so the following returns hashreal unchanged. This is
@ -416,10 +416,10 @@ complex_hash(PyComplexObject *v)
* compare equal must have the same hash value, so that
* hash(x + 0*j) must equal hash(x).
*/
combined = hashreal + 1000003 * hashimag;
if (combined == -1)
combined = -2;
return combined;
combined = hashreal + _PyHASH_IMAG * hashimag;
if (combined == (unsigned long)-1)
combined = (unsigned long)-2;
return (long)combined;
}
/* This macro may return! */

View file

@ -2571,18 +2571,37 @@ long_hash(PyLongObject *v)
sign = -1;
i = -(i);
}
/* The following loop produces a C unsigned long x such that x is
congruent to the absolute value of v modulo ULONG_MAX. The
resulting x is nonzero if and only if v is. */
while (--i >= 0) {
/* Force a native long #-bits (32 or 64) circular shift */
x = (x >> (8*SIZEOF_LONG-PyLong_SHIFT)) | (x << PyLong_SHIFT);
/* Here x is a quantity in the range [0, _PyHASH_MODULUS); we
want to compute x * 2**PyLong_SHIFT + v->ob_digit[i] modulo
_PyHASH_MODULUS.
The computation of x * 2**PyLong_SHIFT % _PyHASH_MODULUS
amounts to a rotation of the bits of x. To see this, write
x * 2**PyLong_SHIFT = y * 2**_PyHASH_BITS + z
where y = x >> (_PyHASH_BITS - PyLong_SHIFT) gives the top
PyLong_SHIFT bits of x (those that are shifted out of the
original _PyHASH_BITS bits, and z = (x << PyLong_SHIFT) &
_PyHASH_MODULUS gives the bottom _PyHASH_BITS - PyLong_SHIFT
bits of x, shifted up. Then since 2**_PyHASH_BITS is
congruent to 1 modulo _PyHASH_MODULUS, y*2**_PyHASH_BITS is
congruent to y modulo _PyHASH_MODULUS. So
x * 2**PyLong_SHIFT = y + z (mod _PyHASH_MODULUS).
The right-hand side is just the result of rotating the
_PyHASH_BITS bits of x left by PyLong_SHIFT places; since
not all _PyHASH_BITS bits of x are 1s, the same is true
after rotation, so 0 <= y+z < _PyHASH_MODULUS and y + z is
the reduction of x*2**PyLong_SHIFT modulo
_PyHASH_MODULUS. */
x = ((x << PyLong_SHIFT) & _PyHASH_MODULUS) |
(x >> (_PyHASH_BITS - PyLong_SHIFT));
x += v->ob_digit[i];
/* If the addition above overflowed we compensate by
incrementing. This preserves the value modulo
ULONG_MAX. */
if (x < v->ob_digit[i])
x++;
if (x >= _PyHASH_MODULUS)
x -= _PyHASH_MODULUS;
}
x = x * sign;
if (x == (unsigned long)-1)

View file

@ -647,63 +647,101 @@ PyObject_RichCompareBool(PyObject *v, PyObject *w, int op)
All the utility functions (_Py_Hash*()) return "-1" to signify an error.
*/
/* For numeric types, the hash of a number x is based on the reduction
of x modulo the prime P = 2**_PyHASH_BITS - 1. It's designed so that
hash(x) == hash(y) whenever x and y are numerically equal, even if
x and y have different types.
A quick summary of the hashing strategy:
(1) First define the 'reduction of x modulo P' for any rational
number x; this is a standard extension of the usual notion of
reduction modulo P for integers. If x == p/q (written in lowest
terms), the reduction is interpreted as the reduction of p times
the inverse of the reduction of q, all modulo P; if q is exactly
divisible by P then define the reduction to be infinity. So we've
got a well-defined map
reduce : { rational numbers } -> { 0, 1, 2, ..., P-1, infinity }.
(2) Now for a rational number x, define hash(x) by:
reduce(x) if x >= 0
-reduce(-x) if x < 0
If the result of the reduction is infinity (this is impossible for
integers, floats and Decimals) then use the predefined hash value
_PyHASH_INF for x >= 0, or -_PyHASH_INF for x < 0, instead.
_PyHASH_INF, -_PyHASH_INF and _PyHASH_NAN are also used for the
hashes of float and Decimal infinities and nans.
A selling point for the above strategy is that it makes it possible
to compute hashes of decimal and binary floating-point numbers
efficiently, even if the exponent of the binary or decimal number
is large. The key point is that
reduce(x * y) == reduce(x) * reduce(y) (modulo _PyHASH_MODULUS)
provided that {reduce(x), reduce(y)} != {0, infinity}. The reduction of a
binary or decimal float is never infinity, since the denominator is a power
of 2 (for binary) or a divisor of a power of 10 (for decimal). So we have,
for nonnegative x,
reduce(x * 2**e) == reduce(x) * reduce(2**e) % _PyHASH_MODULUS
reduce(x * 10**e) == reduce(x) * reduce(10**e) % _PyHASH_MODULUS
and reduce(10**e) can be computed efficiently by the usual modular
exponentiation algorithm. For reduce(2**e) it's even better: since
P is of the form 2**n-1, reduce(2**e) is 2**(e mod n), and multiplication
by 2**(e mod n) modulo 2**n-1 just amounts to a rotation of bits.
*/
long
_Py_HashDouble(double v)
{
double intpart, fractpart;
int expo;
long hipart;
long x; /* the final hash value */
/* This is designed so that Python numbers of different types
* that compare equal hash to the same value; otherwise comparisons
* of mapping keys will turn out weird.
*/
int e, sign;
double m;
unsigned long x, y;
if (!Py_IS_FINITE(v)) {
if (Py_IS_INFINITY(v))
return v < 0 ? -271828 : 314159;
return v > 0 ? _PyHASH_INF : -_PyHASH_INF;
else
return 0;
return _PyHASH_NAN;
}
fractpart = modf(v, &intpart);
if (fractpart == 0.0) {
/* This must return the same hash as an equal int or long. */
if (intpart > LONG_MAX/2 || -intpart > LONG_MAX/2) {
/* Convert to long and use its hash. */
PyObject *plong; /* converted to Python long */
plong = PyLong_FromDouble(v);
if (plong == NULL)
return -1;
x = PyObject_Hash(plong);
Py_DECREF(plong);
return x;
}
/* Fits in a C long == a Python int, so is its own hash. */
x = (long)intpart;
if (x == -1)
x = -2;
return x;
m = frexp(v, &e);
sign = 1;
if (m < 0) {
sign = -1;
m = -m;
}
/* The fractional part is non-zero, so we don't have to worry about
* making this match the hash of some other type.
* Use frexp to get at the bits in the double.
* Since the VAX D double format has 56 mantissa bits, which is the
* most of any double format in use, each of these parts may have as
* many as (but no more than) 56 significant bits.
* So, assuming sizeof(long) >= 4, each part can be broken into two
* longs; frexp and multiplication are used to do that.
* Also, since the Cray double format has 15 exponent bits, which is
* the most of any double format in use, shifting the exponent field
* left by 15 won't overflow a long (again assuming sizeof(long) >= 4).
*/
v = frexp(v, &expo);
v *= 2147483648.0; /* 2**31 */
hipart = (long)v; /* take the top 32 bits */
v = (v - (double)hipart) * 2147483648.0; /* get the next 32 bits */
x = hipart + (long)v + (expo << 15);
if (x == -1)
x = -2;
return x;
/* process 28 bits at a time; this should work well both for binary
and hexadecimal floating point. */
x = 0;
while (m) {
x = ((x << 28) & _PyHASH_MODULUS) | x >> (_PyHASH_BITS - 28);
m *= 268435456.0; /* 2**28 */
e -= 28;
y = (unsigned long)m; /* pull out integer part */
m -= y;
x += y;
if (x >= _PyHASH_MODULUS)
x -= _PyHASH_MODULUS;
}
/* adjust for the exponent; first reduce it modulo _PyHASH_BITS */
e = e >= 0 ? e % _PyHASH_BITS : _PyHASH_BITS-1-((-1-e) % _PyHASH_BITS);
x = ((x << e) & _PyHASH_MODULUS) | x >> (_PyHASH_BITS - e);
x = x * sign;
if (x == (unsigned long)-1)
x = (unsigned long)-2;
return (long)x;
}
long

View file

@ -4921,6 +4921,7 @@ slot_tp_hash(PyObject *self)
PyObject *func, *res;
static PyObject *hash_str;
long h;
int overflow;
func = lookup_method(self, "__hash__", &hash_str);
@ -4937,14 +4938,27 @@ slot_tp_hash(PyObject *self)
Py_DECREF(func);
if (res == NULL)
return -1;
if (PyLong_Check(res))
if (!PyLong_Check(res)) {
PyErr_SetString(PyExc_TypeError,
"__hash__ method should return an integer");
return -1;
}
/* Transform the PyLong `res` to a C long `h`. For an existing
hashable Python object x, hash(x) will always lie within the range
of a C long. Therefore our transformation must preserve values
that already lie within this range, to ensure that if x.__hash__()
returns hash(y) then hash(x) == hash(y). */
h = PyLong_AsLongAndOverflow(res, &overflow);
if (overflow)
/* res was not within the range of a C long, so we're free to
use any sufficiently bit-mixing transformation;
long.__hash__ will do nicely. */
h = PyLong_Type.tp_hash(res);
else
h = PyLong_AsLong(res);
Py_DECREF(res);
if (h == -1 && !PyErr_Occurred())
h = -2;
return h;
if (h == -1 && !PyErr_Occurred())
h = -2;
return h;
}
static PyObject *