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authorMark Dickinson <dickinsm@gmail.com>2010-05-26 16:02:59 (GMT)
committerMark Dickinson <dickinsm@gmail.com>2010-05-26 16:02:59 (GMT)
commit7f1bf8004de5584b42b1d4771caf12e3302ac506 (patch)
treef9339da7d757c4f51af52733f4551f6dca64d65b /Objects/longobject.c
parente73b21240a985583728b40d53f4274bfe406696b (diff)
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Issue #8817: Expose round-to-nearest variant of divmod in _PyLong_Divmod_Near
for use by the datetime module; also refactor long_round to use this function.
Diffstat (limited to 'Objects/longobject.c')
-rw-r--r--Objects/longobject.c251
1 files changed, 140 insertions, 111 deletions
diff --git a/Objects/longobject.c b/Objects/longobject.c
index 564d1a0..3eb0c44 100644
--- a/Objects/longobject.c
+++ b/Objects/longobject.c
@@ -4201,140 +4201,169 @@ long__format__(PyObject *self, PyObject *args)
PyUnicode_GET_SIZE(format_spec));
}
-static PyObject *
-long_round(PyObject *self, PyObject *args)
+/* Return a pair (q, r) such that a = b * q + r, and
+ abs(r) <= abs(b)/2, with equality possible only if q is even.
+ In other words, q == a / b, rounded to the nearest integer using
+ round-half-to-even. */
+
+PyObject *
+_PyLong_Divmod_Near(PyObject *a, PyObject *b)
{
- PyObject *o_ndigits=NULL, *temp;
- PyLongObject *pow=NULL, *q=NULL, *r=NULL, *ndigits=NULL, *one;
- int errcode;
- digit q_mod_4;
-
- /* Notes on the algorithm: to round to the nearest 10**n (n positive),
- the straightforward method is:
-
- (1) divide by 10**n
- (2) round to nearest integer (round to even in case of tie)
- (3) multiply result by 10**n.
-
- But the rounding step involves examining the fractional part of the
- quotient to see whether it's greater than 0.5 or not. Since we
- want to do the whole calculation in integer arithmetic, it's
- simpler to do:
-
- (1) divide by (10**n)/2
- (2) round to nearest multiple of 2 (multiple of 4 in case of tie)
- (3) multiply result by (10**n)/2.
-
- Then all we need to know about the fractional part of the quotient
- arising in step (2) is whether it's zero or not.
-
- Doing both a multiplication and division is wasteful, and is easily
- avoided if we just figure out how much to adjust the original input
- by to do the rounding.
-
- Here's the whole algorithm expressed in Python.
-
- def round(self, ndigits = None):
- """round(int, int) -> int"""
- if ndigits is None or ndigits >= 0:
- return self
- pow = 10**-ndigits >> 1
- q, r = divmod(self, pow)
- self -= r
- if (q & 1 != 0):
- if (q & 2 == r == 0):
- self -= pow
- else:
- self += pow
- return self
+ PyLongObject *quo = NULL, *rem = NULL;
+ PyObject *one = NULL, *twice_rem, *result, *temp;
+ int cmp, quo_is_odd, quo_is_neg;
+
+ /* Equivalent Python code:
+
+ def divmod_near(a, b):
+ q, r = divmod(a, b)
+ # round up if either r / b > 0.5, or r / b == 0.5 and q is odd.
+ # The expression r / b > 0.5 is equivalent to 2 * r > b if b is
+ # positive, 2 * r < b if b negative.
+ greater_than_half = 2*r > b if b > 0 else 2*r < b
+ exactly_half = 2*r == b
+ if greater_than_half or exactly_half and q % 2 == 1:
+ q += 1
+ r -= b
+ return q, r
*/
+ if (!PyLong_Check(a) || !PyLong_Check(b)) {
+ PyErr_SetString(PyExc_TypeError,
+ "non-integer arguments in division");
+ return NULL;
+ }
+
+ /* Do a and b have different signs? If so, quotient is negative. */
+ quo_is_neg = (Py_SIZE(a) < 0) != (Py_SIZE(b) < 0);
+
+ one = PyLong_FromLong(1L);
+ if (one == NULL)
+ return NULL;
+
+ if (long_divrem((PyLongObject*)a, (PyLongObject*)b, &quo, &rem) < 0)
+ goto error;
+
+ /* compare twice the remainder with the divisor, to see
+ if we need to adjust the quotient and remainder */
+ twice_rem = long_lshift((PyObject *)rem, one);
+ if (twice_rem == NULL)
+ goto error;
+ if (quo_is_neg) {
+ temp = long_neg((PyLongObject*)twice_rem);
+ Py_DECREF(twice_rem);
+ twice_rem = temp;
+ if (twice_rem == NULL)
+ goto error;
+ }
+ cmp = long_compare((PyLongObject *)twice_rem, (PyLongObject *)b);
+ Py_DECREF(twice_rem);
+
+ quo_is_odd = Py_SIZE(quo) != 0 && ((quo->ob_digit[0] & 1) != 0);
+ if ((Py_SIZE(b) < 0 ? cmp < 0 : cmp > 0) || (cmp == 0 && quo_is_odd)) {
+ /* fix up quotient */
+ if (quo_is_neg)
+ temp = long_sub(quo, (PyLongObject *)one);
+ else
+ temp = long_add(quo, (PyLongObject *)one);
+ Py_DECREF(quo);
+ quo = (PyLongObject *)temp;
+ if (quo == NULL)
+ goto error;
+ /* and remainder */
+ if (quo_is_neg)
+ temp = long_add(rem, (PyLongObject *)b);
+ else
+ temp = long_sub(rem, (PyLongObject *)b);
+ Py_DECREF(rem);
+ rem = (PyLongObject *)temp;
+ if (rem == NULL)
+ goto error;
+ }
+
+ result = PyTuple_New(2);
+ if (result == NULL)
+ goto error;
+
+ /* PyTuple_SET_ITEM steals references */
+ PyTuple_SET_ITEM(result, 0, (PyObject *)quo);
+ PyTuple_SET_ITEM(result, 1, (PyObject *)rem);
+ Py_DECREF(one);
+ return result;
+
+ error:
+ Py_XDECREF(quo);
+ Py_XDECREF(rem);
+ Py_XDECREF(one);
+ return NULL;
+}
+
+static PyObject *
+long_round(PyObject *self, PyObject *args)
+{
+ PyObject *o_ndigits=NULL, *temp, *result, *ndigits;
+
+ /* To round an integer m to the nearest 10**n (n positive), we make use of
+ * the divmod_near operation, defined by:
+ *
+ * divmod_near(a, b) = (q, r)
+ *
+ * where q is the nearest integer to the quotient a / b (the
+ * nearest even integer in the case of a tie) and r == a - q * b.
+ * Hence q * b = a - r is the nearest multiple of b to a,
+ * preferring even multiples in the case of a tie.
+ *
+ * So the nearest multiple of 10**n to m is:
+ *
+ * m - divmod_near(m, 10**n)[1].
+ */
if (!PyArg_ParseTuple(args, "|O", &o_ndigits))
return NULL;
if (o_ndigits == NULL)
return long_long(self);
- ndigits = (PyLongObject *)PyNumber_Index(o_ndigits);
+ ndigits = PyNumber_Index(o_ndigits);
if (ndigits == NULL)
return NULL;
+ /* if ndigits >= 0 then no rounding is necessary; return self unchanged */
if (Py_SIZE(ndigits) >= 0) {
Py_DECREF(ndigits);
return long_long(self);
}
- Py_INCREF(self); /* to keep refcounting simple */
- /* we now own references to self, ndigits */
-
- /* pow = 10 ** -ndigits >> 1 */
- pow = (PyLongObject *)PyLong_FromLong(10L);
- if (pow == NULL)
- goto error;
- temp = long_neg(ndigits);
+ /* result = self - divmod_near(self, 10 ** -ndigits)[1] */
+ temp = long_neg((PyLongObject*)ndigits);
Py_DECREF(ndigits);
- ndigits = (PyLongObject *)temp;
+ ndigits = temp;
if (ndigits == NULL)
- goto error;
- temp = long_pow((PyObject *)pow, (PyObject *)ndigits, Py_None);
- Py_DECREF(pow);
- pow = (PyLongObject *)temp;
- if (pow == NULL)
- goto error;
- assert(PyLong_Check(pow)); /* check long_pow returned a long */
- one = (PyLongObject *)PyLong_FromLong(1L);
- if (one == NULL)
- goto error;
- temp = long_rshift(pow, one);
- Py_DECREF(one);
- Py_DECREF(pow);
- pow = (PyLongObject *)temp;
- if (pow == NULL)
- goto error;
+ return NULL;
- /* q, r = divmod(self, pow) */
- errcode = l_divmod((PyLongObject *)self, pow, &q, &r);
- if (errcode == -1)
- goto error;
+ result = PyLong_FromLong(10L);
+ if (result == NULL) {
+ Py_DECREF(ndigits);
+ return NULL;
+ }
- /* self -= r */
- temp = long_sub((PyLongObject *)self, r);
- Py_DECREF(self);
- self = temp;
- if (self == NULL)
- goto error;
+ temp = long_pow(result, ndigits, Py_None);
+ Py_DECREF(ndigits);
+ Py_DECREF(result);
+ result = temp;
+ if (result == NULL)
+ return NULL;
- /* get value of quotient modulo 4 */
- if (Py_SIZE(q) == 0)
- q_mod_4 = 0;
- else if (Py_SIZE(q) > 0)
- q_mod_4 = q->ob_digit[0] & 3;
- else
- q_mod_4 = (PyLong_BASE-q->ob_digit[0]) & 3;
+ temp = _PyLong_Divmod_Near(self, result);
+ Py_DECREF(result);
+ result = temp;
+ if (result == NULL)
+ return NULL;
- if ((q_mod_4 & 1) == 1) {
- /* q is odd; round self up or down by adding or subtracting pow */
- if (q_mod_4 == 1 && Py_SIZE(r) == 0)
- temp = (PyObject *)long_sub((PyLongObject *)self, pow);
- else
- temp = (PyObject *)long_add((PyLongObject *)self, pow);
- Py_DECREF(self);
- self = temp;
- if (self == NULL)
- goto error;
- }
- Py_DECREF(q);
- Py_DECREF(r);
- Py_DECREF(pow);
- Py_DECREF(ndigits);
- return self;
+ temp = long_sub((PyLongObject *)self,
+ (PyLongObject *)PyTuple_GET_ITEM(result, 1));
+ Py_DECREF(result);
+ result = temp;
- error:
- Py_XDECREF(q);
- Py_XDECREF(r);
- Py_XDECREF(pow);
- Py_XDECREF(self);
- Py_XDECREF(ndigits);
- return NULL;
+ return result;
}
static PyObject *