我有一个非常简单的程序,它有一个作为基类实现的单例,它简单地定义了一个用于实现“单一性”的 new。但是,我无法再向 init 方法发送参数 - 当我这样做时,我从单例 new super 调用中收到“TypeError: object() takes no parameters”:
class Singleton(object):
_instances = {}
def __new__(cls, *args, **kwargs):
print(args, kwargs)
if cls._instances.get(cls, None) is None:
cls._instances[cls] = super(Singleton, cls).__new__(cls, *args, **kwargs)
return Singleton._instances[cls]
class OneOfAKind(Singleton):
def __init__(self):
print('--> OneOfAKind __init__')
Singleton.__init__(self)
class OneOfAKind2(Singleton):
def __init__(self, onearg):
print('--> OneOfAKind2 __init__')
Singleton.__init__(self)
self._onearg = onearg
x = OneOfAKind()
y = OneOfAKind()
print(x == y)
X = OneOfAKind2('testing')
输出是:
() {}
Traceback (most recent call last):
File "./mytest.py", line 29, in <module>
x = OneOfAKind()
File "./mytest.py", line 10, in __new__
cls._instances[cls] = super(Singleton, cls).__new__(cls, (), {})
TypeError: object() takes no parameters
最佳答案
是的,因为 object
是您从 super
继承和调用的对象,它不带任何参数:
In [54]: object('foo', 'bar')
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-54-f03d0963548d> in <module>()
----> 1 object('foo', 'bar')
TypeError: object() takes no parameters
如果你想做这样的事情,我推荐一个 metaclass
而不是子类化,而不是覆盖 __new__
metaclass
覆盖 __call__
用于对象创建:
import six ## used for compatibility between py2 and py3
class Singleton(type):
_instances = {}
def __call__(cls, *args, **kwargs):
if Singleton._instances.get(cls, None) is None:
Singleton._instances[cls] = super(Singleton, cls).__call__(*args, **kwargs)
return Singleton._instances[cls]
@six.add_metaclass(Singleton)
class OneOfAKind(object):
def __init__(self):
print('--> OneOfAKind __init__')
@six.add_metaclass(Singleton)
class OneOfAKind2(object):
def __init__(self, onearg):
print('--> OneOfAKind2 __init__')
self._onearg = onearg
然后:
In [64]: OneOfAKind() == OneOfAKind()
--> OneOfAKind __init__
Out[64]: True
In [65]: OneOfAKind() == OneOfAKind()
Out[65]: True
In [66]: OneOfAKind() == OneOfAKind2('foo')
Out[66]: False
关于python 覆盖 __new__ 无法将参数发送到 __init__,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48039587/