我尝试了以下实验:
>>> class A(object):
... def __init__(self, name):
... self.name = name
... def __str__(self):
... return self.name
... def __eq__(self, other):
... print('{}.__eq__({})'.format(self, other))
... return NotImplemented
...
>>> a1 = A('a1')
>>> a2 = A('a2')
>>> a1 == a2
a1.__eq__(a2)
a2.__eq__(a1)
a1.__eq__(a2)
a2.__eq__(a1)
a2.__eq__(a1)
a1.__eq__(a2)
False
这里到底发生了什么?
更一般而言,是否存在评估操作员时发生的确切流程的官方文档? data model文档暗示了某种后备行为,但未精确描述其含义。有几种使情况复杂化的可能性:
a
和b
可以是相同或不同的类型a
和b
可能会或可能不会定义__eq__
方法a.__eq__(b)
或b.__eq__(a)
可能返回NotImplemented
或其他值。某种流程图会有所帮助。
编辑:在将此问题标记为重复之前,请确保假定的重复回答以下问题:
为什么
__eq__
在给定模式下被调用6次?该行为在何处得到充分记录?
我可以得到流程图吗?
最佳答案
该行为仅是python-2.x,并且是内部比较工作的方式(至少是CPython)的一部分,但前提是两者都是新样式类且两个参数都具有相同的类型!
源C代码读取(我突出显示了完成和/或跳过比较的部分):
PyObject *
PyObject_RichCompare(PyObject *v, PyObject *w, int op)
{
PyObject *res;
assert(Py_LT <= op && op <= Py_GE);
if (Py_EnterRecursiveCall(" in cmp"))
return NULL;
/* If the types are equal, and not old-style instances, try to
get out cheap (don't bother with coercions etc.). */
if (v->ob_type == w->ob_type && !PyInstance_Check(v)) {
cmpfunc fcmp;
richcmpfunc frich = RICHCOMPARE(v->ob_type);
/* If the type has richcmp, try it first. try_rich_compare
tries it two-sided, which is not needed since we've a
single type only. */
if (frich != NULL) {
/****************************************************/
/* 1. This first tries v.__eq__(w) then w.__eq__(v) */
/****************************************************/
res = (*frich)(v, w, op);
if (res != Py_NotImplemented)
goto Done;
Py_DECREF(res);
}
/* No richcmp, or this particular richmp not implemented.
Try 3-way cmp. */
fcmp = v->ob_type->tp_compare;
if (fcmp != NULL)
/***********************************************/
/* Skipped because you don't implement __cmp__ */
/***********************************************/
int c = (*fcmp)(v, w);
c = adjust_tp_compare(c);
if (c == -2) {
res = NULL;
goto Done;
}
res = convert_3way_to_object(op, c);
goto Done;
}
}
/* Fast path not taken, or couldn't deliver a useful result. */
res = do_richcmp(v, w, op);
Done:
Py_LeaveRecursiveCall();
return res;
}
/* Try a genuine rich comparison, returning an object. Return:
NULL for exception;
NotImplemented if this particular rich comparison is not implemented or
undefined;
some object not equal to NotImplemented if it is implemented
(this latter object may not be a Boolean).
*/
static PyObject *
try_rich_compare(PyObject *v, PyObject *w, int op)
{
richcmpfunc f;
PyObject *res;
if (v->ob_type != w->ob_type &&
PyType_IsSubtype(w->ob_type, v->ob_type) &&
(f = RICHCOMPARE(w->ob_type)) != NULL) {
/*******************************************************************************/
/* Skipped because you don't compare unequal classes where w is a subtype of v */
/*******************************************************************************/
res = (*f)(w, v, _Py_SwappedOp[op]);
if (res != Py_NotImplemented)
return res;
Py_DECREF(res);
}
if ((f = RICHCOMPARE(v->ob_type)) != NULL) {
/*****************************************************************/
/** 2. This again tries to evaluate v.__eq__(w) then w.__eq__(v) */
/*****************************************************************/
res = (*f)(v, w, op);
if (res != Py_NotImplemented)
return res;
Py_DECREF(res);
}
if ((f = RICHCOMPARE(w->ob_type)) != NULL) {
/***********************************************************************/
/* 3. This tries the reversed comparison: w.__eq__(v) then v.__eq__(w) */
/***********************************************************************/
return (*f)(w, v, _Py_SwappedOp[op]);
}
res = Py_NotImplemented;
Py_INCREF(res);
return res;
}
有趣的部分是评论-可以回答您的问题:
如果两者都是相同的类型和新样式的类,则假定它可以做一个捷径:它尝试进行丰富的比较。正常返回和反向返回NotImplemented并继续执行。
它进入
try_rich_compare
函数,在那里尝试再次比较它们,首先是正常值,然后是相反值。通过测试反向操作进行最后一次尝试:现在,它比较反向操作,然后再次尝试正常(反向操作的反向操作)。
(未显示)最后,所有3种可能性均失败,然后如果对象相同的
a1 is a2
进行最后测试,则返回观察到的False
。如果测试
a1 == a1
,则可以观察到最后一个测试的存在:>>> a1 == a1
a1.__eq__(a1)
a1.__eq__(a1)
a1.__eq__(a1)
a1.__eq__(a1)
a1.__eq__(a1)
a1.__eq__(a1)
True
我不知道该行为是否已得到充分记录,至少在
__eq__
的文档中有一些提示如果富比较方法未实现给定参数对的操作,则它可能返回单例NotImplemented。
和
__cmp__
:如果未定义丰富比较(请参见上文),则由比较操作调用。
其他一些观察:
请注意,如果您定义
__cmp__
,则它不像return NotImplemented
那样尊重__eq__
(因为它在PyObject_RichCompare
中输入了先前跳过的分支):class A(object):
def __init__(self, name):
self.name = name
def __str__(self):
return self.name
def __eq__(self, other):
print('{}.__eq__({})'.format(self, other))
return NotImplemented
def __cmp__(self, other):
print('{}.__cmp__({})'.format(self, other))
return NotImplemented
>>> a1, a2 = A('a1'), A('a2')
>>> a1 == a2
a1.__eq__(a2)
a2.__eq__(a1)
a1.__cmp__(a2)
a2.__cmp__(a1)
False
如果您显式地与超类和继承的类进行比较,则可以轻松看出子类或相同类的行为:
>>> class A(object):
... def __init__(self, name):
... self.name = name
... def __str__(self):
... return self.name
... def __eq__(self, other):
... print('{}.__eq__({}) from A'.format(self, other))
... return NotImplemented
...
>>>
>>> class B(A):
... def __eq__(self, other):
... print('{}.__eq__({}) from B'.format(self, other))
... return NotImplemented
...
>>>
>>> a1, a2 = A('a1'), B('a2')
>>> a1 == a2
a2.__eq__(a1) from B
a1.__eq__(a2) from A
a1.__eq__(a2) from A
a2.__eq__(a1) from B
a2.__eq__(a1) from B
a1.__eq__(a2) from A
False
>>> a2 == a1
a2.__eq__(a1) from B
a1.__eq__(a2) from A
a1.__eq__(a2) from A
a2.__eq__(a1) from B
False
最后的评论:
我添加了用于“打印”的代码,它在gist中进行比较。如果您知道如何创建python-c-extensions,则可以自己编译和运行代码(需要使用两个参数调用
myrichcmp
函数以进行相等性比较)。