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

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(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
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26
Objects/typeobject.c
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@ -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 *