【发布时间】:2014-06-04 22:41:12
【问题描述】:
这个示例代码有点奇怪,但请耐心等待......
class Foo(object):
def __init__(self, internal_dict = None):
self._internal_dict = internal_dict or {}
for attribute_name in self.__class__.__dict__.keys():
attr = getattr(self.__class__, attribute_name)
if isinstance(attr, str) and attribute_name.startswith("a"):
# We are iterating over all string attributes of this class whos name begins with "a"
self._internal_dict[attribute_name] = {}
setattr(self, attribute_name + '_nested_object', Foo(internal_dict=self._internal_dict[attribute_name]))
class FooChild(Foo):
ax = "5"
ay = "10"
fc = FooChild()
print fc.ax_nested_object._internal_dict # This prints {}
fc.ax_nested_object._internal_dict['123'] = 'abc'
print fc._internal_dict # This prints {'ay': {}, 'ax': {}}
我本来希望我的{'123' = 'abc'} 能够通过第二次打印,因为字典应该通过引用传递到递归__init__ 调用中。但是,如果我更改此行:
self._internal_dict[attribute_name] = {}
到这里:
self._internal_dict[attribute_name] = {'test': 1}
然后我得到以下打印:
{'test': 1}
{'ay': {'test': 1}, 'ax': {'test': 1, '123': 'abc'}}
为什么启动该字典数据会导致它通过引用正确传递?
【问题讨论】:
-
这就是我让人们在检查是否提供可选参数时询问是否包含
is None的确切原因。感谢您确认这确实发生了 ;-)