def __index__(self):
return self.value
+# int subclass with broken arithmetic operators; implementations must
+# convert their arguments to exact ints instead of using these.
+class BadIntSubclass(int):
+ def _binop(self, other='ignored', mod=None):
+ return 42
+ __add__ = __radd__ = __sub__ = __rsub__ = _binop
+ __mul__ = __rmul__ = __mod__ = __rmod__ = _binop
+ __divmod__ = __rdivmod__ = __pow__ = __rpow__ = _binop
+ __floordiv__ = __rfloordiv__ = _binop
+ __lshift__ = __rlshift__ = __rshift__ = __rrshift__ = _binop
+ __and__ = __rand__ = __or__ = __ror__ = __xor__ = __rxor__ = _binop
+ __lt__ = __le__ = __gt__ = __ge__ = _binop
+
# Here's a pure Python version of the math.integer.factorial algorithm, for
# documentation and comparison purposes.
#
self.assertIntEqual(isqrt(False), 0)
self.assertIntEqual(isqrt(MyIndexable(1729)), 41)
+ # Overridden operators of an int subclass must not affect the
+ # result.
+ self.assertIntEqual(isqrt(BadIntSubclass(10**20)), 10**10)
+ self.assertIntEqual(isqrt(BadIntSubclass(10**20 - 1)), 10**10 - 1)
+
with self.assertRaises(ValueError):
isqrt(MyIndexable(-3))
/* The correct result is either a or a - 1. Figure out which, and
decrement a if necessary. */
- /* a_too_large = n < a * a */
+ /* a_too_large = n < a * a. Compare by value: n can be an instance
+ of an int subclass with an overridden __lt__ method. */
b = PyNumber_Multiply(a, a);
if (b == NULL) {
goto error;
}
- a_too_large = PyObject_RichCompareBool(n, b, Py_LT);
+ PyObject *cmp = PyLong_Type.tp_richcompare(n, b, Py_LT);
Py_DECREF(b);
- if (a_too_large == -1) {
+ if (cmp == NULL) {
goto error;
}
+ assert(PyBool_Check(cmp));
+ a_too_large = (cmp == Py_True);
+ Py_DECREF(cmp);
if (a_too_large) {
Py_SETREF(a, PyNumber_Subtract(a, _PyLong_GetOne()));